Math Jams

## State Competition Review Math Jam with Richard Rusczyk

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As part of MATHCOUNTS Week, Art of Problem Solving’s Richard Rusczyk and David Patrick will discuss how to solve some of the toughest problems from the 2020 State MATHCOUNTS competition.

#### Facilitator: Richard Rusczyk

rrusczyk 2020-05-14 15:14:08
It's time to get started!
rrusczyk 2020-05-14 15:14:17
I'm Richard Rusczyk. You might recognize me from the MATHCOUNTS Minis. My first MATHCOUNTS event was over 34 years ago. Yes, MATHCOUNTS really is that old.
rrusczyk 2020-05-14 15:14:19
(And, yes, that means I'm even older.)
rrusczyk 2020-05-14 15:14:34
The following year I made it to Nationals, representing Alabama.
rrusczyk 2020-05-14 15:14:42
Since then, I've returned to Nationals around 20 times. For the last 12-15 years, my colleague tonight (David Patrick, DPatrick) and I spent the entire Registration Day at the registration desk visiting with students and old friends. If you have AoPS books in your home, you probably recognize DPatrick's name from some of the covers!
DPatrick 2020-05-14 15:14:45
Hi!
RP3.1415 2020-05-14 15:14:57
yay!!
Themiihacker 2020-05-14 15:14:57
hey
menlo 2020-05-14 15:14:57
Hi!
yekolo 2020-05-14 15:14:57
hi!
SandyBeach11 2020-05-14 15:14:57
Hi!
genius_007 2020-05-14 15:14:57
Hello!
bingo2016 2020-05-14 15:14:57
Hello!
floatmeeting 2020-05-14 15:14:57
HI
bobjoebilly 2020-05-14 15:14:57
Hi
MathWizard09 2020-05-14 15:14:57
hi!
math_piggy 2020-05-14 15:14:57
Hi!
motao 2020-05-14 15:14:57
Hello
Abhinac 2020-05-14 15:14:57
hi
LJCoder619 2020-05-14 15:14:57
Hi!
rrusczyk 2020-05-14 15:15:12
We couldn't go to MATHCOUNTS this year. We were going to miss you all so much that we partnered with MATHCOUNTS to bring MATHCOUNTS to us.
PI-EATEN 2020-05-14 15:15:25
Yay
SparklyFlowers 2020-05-14 15:15:25
Yay! I'm so excited.
LANSH 2020-05-14 15:15:25
yay!
sdattilo2002 2020-05-14 15:15:25
YAY!!!!!
superls 2020-05-14 15:15:25
yay!
guo35140 2020-05-14 15:15:25
3 14!!!!!
Piano_Man123 2020-05-14 15:15:25
Yay!
Layla2018 2020-05-14 15:15:25
YAY!!
Mathematician1010 2020-05-14 15:15:32
yay!
Kesav 2020-05-14 15:15:32
awesome
KLBBC 2020-05-14 15:15:32
Yeah
jellybeanchocolate 2020-05-14 15:15:37
yay!
bedwinprusik578 2020-05-14 15:15:37
yay
INDYMATHz 2020-05-14 15:15:37
nice
LAC08512 2020-05-14 15:15:37
YAY!
rrusczyk 2020-05-14 15:15:48
We have two special guests this afternoon from MATHCOUNTS. I've had the great fortune to work with MATHCOUNTS for even longer than this website has existed, and I've always been amazed at what they pull off each year.
rrusczyk 2020-05-14 15:16:04
Many of you here today know about the Competition Series, but you might not have thought of how much work goes into making contests happen. Or the fact that it needs to happen around 600 times. Staffed almost entirely with volunteers. It still blows my mind that they've been able to make that work for nearly 40 years now.
rrusczyk 2020-05-14 15:16:13
On top of that, they keep adding more, like the Club Program and the Math Video Challenge that many of you have been part of in the past few years.
rrusczyk 2020-05-14 15:16:24
Representing MATHCOUNTS today, we have Executive Director Kristen Chandler (LadyBlue32) and Senior Manager of Education Kera Johnson (KDEJ22). If you took the test earlier this week, you saw Kristen and her indoor pool in the intro video before the test.
mathw0nder 2020-05-14 15:17:24
Hi Kristen and Kera!
bestzack66 2020-05-14 15:17:24
Hello!
floatmeeting 2020-05-14 15:17:24
hi!
Captus 2020-05-14 15:17:24
Hi!
sixoneeight 2020-05-14 15:17:24
Hey!
OlympusHero 2020-05-14 15:17:24
Wow, this is awesome! Hi Kristen and Kera!
hi!
CSPAL 2020-05-14 15:17:24
Hello!!!
hi
zhuqingzhang 2020-05-14 15:17:24
hi
kdej22 2020-05-14 15:17:30
hello!
kdej22 2020-05-14 15:18:10
rrusczyk 2020-05-14 15:18:30
Next let's talk about how the classroom works.
Hey everyone! We're thrilled to be joining you!! Thanks for making MATHCOUNTS Week so successful@
nnnF1s 2020-05-14 15:19:04
Thank you
Hard 2020-05-14 15:19:04
Thank you! I was awesome!
tkfun 2020-05-14 15:19:04
hello! thanks!
Noam2007 2020-05-14 15:19:04
Thank you!
Cozzmo 2020-05-14 15:19:09
Yay! thx 4 being here
ObjectZ 2020-05-14 15:19:10
You're welcome!
MathJams 2020-05-14 15:19:14
Thanks for doing this
jabuticaba149 2020-05-14 15:19:22
YAY!!
Sports8531 2020-05-14 15:19:22
Thanks
s0arbeacon 2020-05-14 15:19:22
No, Thank yOU!
fasterthanlight 2020-05-14 15:19:22
It's an honor to meet you!
rrusczyk 2020-05-14 15:19:26
As you've probably figured out by now, the classroom is moderated, meaning that students can type into the classroom, but these comments will not go directly into the room. These comments go to the instructors, who may choose to share your comments with the room.
rrusczyk 2020-05-14 15:19:36
This helps keep the class organized and on track. This also means that only well-written comments will be dropped into the classroom, so please take time writing responses that are complete and easy to read.
rrusczyk 2020-05-14 15:19:42
You can adjust the screen sizing in a variety of ways; give it a try! Also, you can change the font size by clicking the "A"s at the top of the room. Every few years I have to click the big A one more time, sigh.
rrusczyk 2020-05-14 15:19:56
There are bunches and bunches of students here. As I said, only a fraction of the well-written comments will be passed to the entire group. Please do not take it personally if your comments do not get posted, and please do not complain about it. I expect this Math Jam to be much larger than our typical class, so please be patient with me---there are quite a few of you here today!
rrusczyk 2020-05-14 15:20:11
For those keeping score at home, we have over 1100 people in the room right now.
rrusczyk 2020-05-14 15:20:17
In our regular classes, there are (far, far) fewer students, and everyone gets plenty of "air time" if they participate.
rrusczyk 2020-05-14 15:20:26
Also, we won't be going through the math quite as thoroughly as we do in our classes -- we can't teach all the prerequisite material for every problem as we go. Another difference between today and our regular online classes is that it is very unlikely that we'll be able to answer every single question you ask. We usually do in our classes, but we have a very, very large number of students today!
rrusczyk 2020-05-14 15:20:42
We do have three more special people to introduce.
rrusczyk 2020-05-14 15:20:45
Helping us here today, we have three assistants: Zach Stein-Perlman (ZachSteinPerlman), Agustin Marchionna (tuvie), and Bogdan Blaga (nasrl).
rrusczyk 2020-05-14 15:20:51
Zach has loved mathematics, brainteasers, and logic puzzles for longer than he can remember. He joined AoPS in 2011, back in the days when you faxed in homework. In addition to math, Zach is interested in philosophy and government. In his free time, Zach likes reading fantasy novels, cuddling with his cats, and running half marathons.
ZachSteinPerlman 2020-05-14 15:20:56
Hi!
AOPSmathematics 2020-05-14 15:21:07
Hi!!!
EvanZ 2020-05-14 15:21:07
hello!
floatmeeting 2020-05-14 15:21:07
Hi!
mathcounting 2020-05-14 15:21:07
hi
memi 2020-05-14 15:21:07
hi!!!!
jupiter314 2020-05-14 15:21:07
hello
Heavytoothpaste 2020-05-14 15:21:07
hi
HI
s979903 2020-05-14 15:21:07
Hello!
hi
GrantStar 2020-05-14 15:21:07
hello!
Chesssaga 2020-05-14 15:21:07
Hi!
ObjectZ 2020-05-14 15:21:07
Howdy!
G1-100 2020-05-14 15:21:07
Hi
Ilikeminecraft 2020-05-14 15:21:07
hi
falcon22 2020-05-14 15:21:07
hi!
Stunji_Ray 2020-05-14 15:21:07
hi
OlympusHero 2020-05-14 15:21:15
Hi Zach!
PureSwag 2020-05-14 15:21:15
Hi Zach!!!
math_piggy 2020-05-14 15:21:15
Hi Zach!
Dumdum990 2020-05-14 15:21:15
Hello Zach!
aaravdodhia 2020-05-14 15:21:15
Hey Zach!
rrusczyk 2020-05-14 15:21:17
Agustin joined AoPS recently in 2019. He discovered his passion for maths when he first attended a math olympiad in 5th grade. Since then, he participated in many math olympiads, obtaining several prizes, such as a bronze medal in the IMO. After graduating from high school, he developed a new passion for programming contests. He is attending the University of Buenos Aires for a BS and MS in maths and computer science. He now helps in the organization of the Argentinian Math Olympiad, including tutoring and grading, giving back what he has been given during his time as olympian. In his spare time, Agustin loves to play soccer, as well as soccer related video games.
tuvie 2020-05-14 15:21:23
Hi everyone!
douglasubella 2020-05-14 15:21:39
Hi Agustin!!
Possible 2020-05-14 15:21:39
Hey Agustin!
MathWizard09 2020-05-14 15:21:39
Hi Agustin!
DaBobWhoLikeMath1234 2020-05-14 15:21:39
Hi Agustin!
CosmoMonkhouse 2020-05-14 15:21:41
Hi Agustin!
Speedstorm 2020-05-14 15:21:43
Hi Agustin.
akmathisfun2020 2020-05-14 15:21:43
Hi Agustin!!!
rrusczyk 2020-05-14 15:21:47
Bogdan is finishing his second year at Oxford University, studying Mathematics and Computer Science. He's been passionate about mathematics ever since his grandpa showed him the multiplication so that he can help with keeping scores at card games. He plans on pursuing a PhD, probably in pure math, but there are so many cool things to study! Outside of school, he really enjoys playing video games, basketball, and buying as many colorful socks as possible.
Mathworker2705 2020-05-14 15:22:12
Hi Bogdan!!!
KNM 2020-05-14 15:22:12
Hello Bogdan!
vanigarcha 2020-05-14 15:22:12
Hey Bogdan!
MathWizard38025 2020-05-14 15:22:12
Hey Bogdan!
Mandakini 2020-05-14 15:22:12
Hiiii
CHIPPER33 2020-05-14 15:22:12
Bogdon!!
Handiccraft 2020-05-14 15:22:12
hi bogdan!
Mathematician1010 2020-05-14 15:22:12
Hi Bogdan!
Elaine_Wang 2020-05-14 15:22:12
Hi!
goodskate 2020-05-14 15:22:12
Hi Bogdan!
Piano_Man123 2020-05-14 15:22:12
HI BOgdan!!!!!
areevv 2020-05-14 15:22:12
hi bogdan!!
YuryK 2020-05-14 15:22:17
Hi Bogdan!
IceNeptune 2020-05-14 15:22:17
Hey Bogdan!!!
Facejo 2020-05-14 15:22:17
HELLO BOGDAN!!
rrusczyk 2020-05-14 15:22:31
So, please go ahead and ask questions, but also please understand if we aren't able to answer them all! Our star assistants will get to as many as they can!
rrusczyk 2020-05-14 15:22:39
Today we're going to work through some of the hardest problems on this week's competition. I know I won't be able to keep up with you through the whole Math Jam, so DPatrick is going to join me in working through the problems with all of you.
Handiccraft 2020-05-14 15:22:52
can we start?
rrusczyk 2020-05-14 15:22:56
Good idea!
rrusczyk 2020-05-14 15:23:02
We'll start with the final six problems from the Sprint Round. No calculators allowed!
rrusczyk 2020-05-14 15:23:05
Sprint #25: Equilateral triangle $ABC$ with side length $11$ is constructed on a triangular lattice grid with $12$ lattice points on each side. One of the lattice points inside the triangle is randomly selected and labeled point $P.$ If $a, b$ and $c$ are the distances from $P$ to sides $BC, AC,$ and $AB,$ respectively, as shown, what is the probability that a triangle exists with side lengths $a, b,$ and $c?$ Express your answer as a common fraction.
rrusczyk 2020-05-14 15:23:13
rrusczyk 2020-05-14 15:23:26
Note that you can adjust the bar between the top section and the main room.
rrusczyk 2020-05-14 15:23:45
You can also double-click on images to pop them out into their own windows.
arirah9 2020-05-14 15:23:48
this was actually quite hard
rrusczyk 2020-05-14 15:24:10
A lot of these problems were quite hard. My theory is that the test writers forgot that Luke and Andrew aren't in middle school anymore
romani 2020-05-14 15:24:31
How do you solve it?
rrusczyk 2020-05-14 15:24:35
Great question. How do we usually tackle probability problems like this?
IceNeptune 2020-05-14 15:25:10
you need desired/total
menlo 2020-05-14 15:25:10
success/all possibilites
falcon22 2020-05-14 15:25:21
Successful Outcomes/Total Possible Outcomes
sv739 2020-05-14 15:25:23
successful outcomes/ all possibilites
rrusczyk 2020-05-14 15:25:26
We need to compute $\dfrac{\text{number of points that work}}{\text{total number of points}}$.
rrusczyk 2020-05-14 15:25:40
This should work because the points are equally likely to be chosen.
rrusczyk 2020-05-14 15:25:47
Let's start with the denominator. How many total interior lattice points are there to choose from?
CT17 2020-05-14 15:26:27
There are $\frac{9\cdot 8}{2} = 45$ total points.
CHIPPER33 2020-05-14 15:26:27
45
fasterthanlight 2020-05-14 15:26:27
45
walrus987 2020-05-14 15:26:27
45
KevinW12345 2020-05-14 15:26:27
45
MTHJJS 2020-05-14 15:26:27
1+2+3+4+5+6+7+8+9=45
WAWATHEGOAT 2020-05-14 15:26:27
9+8+7+6... = 45
Mrs. H math teacher 2020-05-14 15:26:27
1 + 2 + 3 + . . . + 9
rrusczyk 2020-05-14 15:26:52
I see a lot of you answered 78 here -- have to be careful!
rrusczyk 2020-05-14 15:27:04
We are choosing from the points inside the triangle!
rrusczyk 2020-05-14 15:27:10
There's $1$ point in the top row (nearest $A$), then $2$ points in the next row, and so on.
rrusczyk 2020-05-14 15:27:14
skyleristhecoolest 2020-05-14 15:27:19
the sides don't count
jacobwu 2020-05-14 15:27:19
the edge doesn't count
rrusczyk 2020-05-14 15:27:25
Exactly!
rrusczyk 2020-05-14 15:27:29
Adding the number of lattice points inside $\triangle ABC$ in each horizontal row, there are $1+2+3+4+5+6+7+8+9=45$lattice points from which we can choose $P.$
rrusczyk 2020-05-14 15:27:34
Next, we count the number of points $P$ that work. Those for which the distances $a,b,c$ could possibly be the side lengths of a triangle.
rrusczyk 2020-05-14 15:27:41
This is where it gets tricky!
rrusczyk 2020-05-14 15:27:46
How might we think about the distances $a,b,c?$
RisingRook 2020-05-14 15:28:10
trinagle inquality
eibc 2020-05-14 15:28:10
triangle innequality!
menlo 2020-05-14 15:28:10
use triangle inequality
jacobwu 2020-05-14 15:28:10
triangle inequality
Windigo 2020-05-14 15:28:10
We can use triangle inequalities
Arnav.Dagar.sd 2020-05-14 15:28:10
Triangles inequality
Mathemagician10 2020-05-14 15:28:10
a + b > c
rrusczyk 2020-05-14 15:28:28
Indeed, we'll need $a+b>c$ and so on from the Triangle Inequality.
rrusczyk 2020-05-14 15:28:46
But we're going to need to learn a little more about $a$, $b$, and $c$.
MathJams 2020-05-14 15:28:52
perpendicular distances?
rrusczyk 2020-05-14 15:29:08
Perpendicular distances. Hmmmm.... What might that make us think of?
amuthup 2020-05-14 15:29:19
heights of triangles ABP, BCP, CAP
mathgirl199 2020-05-14 15:29:26
triangle heights.
shalissa 2020-05-14 15:29:29
height of a triangle
MathWizard38025 2020-05-14 15:29:35
Heights
PShucks 2020-05-14 15:29:35
heights
douglasubella 2020-05-14 15:29:39
Heights!
sixoneeight 2020-05-14 15:29:39
The heihts
AbhiWwis 2020-05-14 15:29:39
altitudes
ZippityA 2020-05-14 15:29:39
altitudes
mathisfun17 2020-05-14 15:29:39
altitudes
rrusczyk 2020-05-14 15:29:44
We notice that our distances from $P$ to the sides of $\triangle ABC$ are altitudes of triangles $PAB$, $PBC$, and $PCA$.
rrusczyk 2020-05-14 15:29:48
rayfish 2020-05-14 15:30:09
sum of heights must be the same
rrusczyk 2020-05-14 15:30:16
Bold claim!
DreamShake34 2020-05-14 15:30:28
i it right?
Facejo 2020-05-14 15:30:33
How do we prove?
rrusczyk 2020-05-14 15:30:39
Let's see!
rrusczyk 2020-05-14 15:30:57
What do altitudes make us think of?
amuthupss 2020-05-14 15:31:14
Areas
undefined_ 2020-05-14 15:31:14
area
Math5K 2020-05-14 15:31:14
area
Kevin06 2020-05-14 15:31:14
Area!
Leonard_my_dude 2020-05-14 15:31:14
areas
Chesssaga 2020-05-14 15:31:14
Area
CoolCarsOnTheRun 2020-05-14 15:31:14
areas
green_arrow 2020-05-14 15:31:14
area
RankDoughnut255 2020-05-14 15:31:14
area
Wesman 2020-05-14 15:31:14
Area
ryanfu2008 2020-05-14 15:31:14
area
rrusczyk 2020-05-14 15:31:28
Great; let's talk area.
Leonard_my_dude 2020-05-14 15:31:51
sum of areas is area of entire triangle so sum of heights is equal to total height
IceWolf10 2020-05-14 15:31:51
they have to add up to the same area
AlphaBetaGammaOmega 2020-05-14 15:31:51
The total area is constant
Noam2007 2020-05-14 15:31:51
$[APC]+[BPC]+[CPA]=[ABC].$
I_like_elmo 2020-05-14 15:31:57
ohh, the areas of the traingle will always equal the big traingle, and the bases are the same, therefore, the altidudes have the same sum
rrusczyk 2020-05-14 15:32:09
Very pretty. Let's get through this step by step.
rrusczyk 2020-05-14 15:32:13
Letting $[ABC]$ represent the area of triangle $ABC,$ we can split up $ABC$ as $$[ABC] = [PAB] + [PBC] + [PCA].$$
rrusczyk 2020-05-14 15:32:23
And how can we write the areas of those smaller triangles?
amuthup 2020-05-14 15:33:17
side length * a/2, side length * b/2, side length * c/2
rocketsri 2020-05-14 15:33:17
11a/2+11b/2+11c/2
Meyzeek_Saveer 2020-05-14 15:33:17
11a +11b+11c all over 2
greenmathcounts 2020-05-14 15:33:17
11a/2+11b/2+11c/2
AWaltz 2020-05-14 15:33:17
11a/2, 11b/2, and 11c/2
pvmathnerd 2020-05-14 15:33:17
11a/2+11b/2+11c/2
emerald_block 2020-05-14 15:33:17
$\frac{as}{2}, \frac{bs}{2}, \frac{bs}{2}$ where $s$ is the side length
rrusczyk 2020-05-14 15:33:26
We use the sides of $ABC$ and the altitudes that we know: \begin{align*}[ABC] &= [PAB]+[PBC]+[PCA]\\

&= \frac{(AB)(c)}{2} + \frac{(BC)(a)}{2} + \frac{(CA)(b)}{2}.\end{align*}
rrusczyk 2020-05-14 15:33:34
We know that $AB = AC = BC = 11$. (Pretty convenient that triangle $ABC$ is equilateral!)
rrusczyk 2020-05-14 15:33:38
So this simplifies to just $[ABC] = \dfrac{11}{2}(a+b+c).$
rrusczyk 2020-05-14 15:33:44
Um, how does that help?
snorlax186 2020-05-14 15:34:15
a+b+c is constant
yizhou76 2020-05-14 15:34:15
a+b+c is constant
christopherp 2020-05-14 15:34:15
a+b+c must be constant
MR_67 2020-05-14 15:34:15
You know the sum of a, b, and c
Zoobat 2020-05-14 15:34:15
a+b+c is constant
A395134 2020-05-14 15:34:15
$a+b+c$ is constant.
eibc 2020-05-14 15:34:15
now we know that a+b+c = the height
rrusczyk 2020-05-14 15:34:35
The area of $ABC$ is constant, so $a+b+c$ must be constant.
rrusczyk 2020-05-14 15:34:40
Specifically, we note that if $h$ is the height of $\triangle ABC,$ we have $[ABC] = \dfrac{1}{2}(BC)(h) = \dfrac{11}{2}(h).$
MathWizard10 2020-05-14 15:34:59
a+b+c=height
jason543 2020-05-14 15:35:05
so h =a+b+c
Tony_Li2007 2020-05-14 15:35:07
a+b+c is height
MR_67 2020-05-14 15:35:13
a+b+c is always height
rrusczyk 2020-05-14 15:35:16
So $a+b+c = h,$ the height of $ABC.$
dc495 2020-05-14 15:35:31
yay we proved it!
rrusczyk 2020-05-14 15:35:37
We have indeed!!!
yukrant1 2020-05-14 15:35:44
This is Viviani’s theorem
v4913 2020-05-14 15:35:44
That's Viviani's theorem
rrusczyk 2020-05-14 15:35:50
I'm going to take your word for it
rrusczyk 2020-05-14 15:35:58
Now what?
turtles0120 2020-05-14 15:36:09
but we're not done?
rrusczyk 2020-05-14 15:36:35
Correct -- we aren't there yet. We need to figure out how many points we can choose that will satisfy our condition in the problem.
rrusczyk 2020-05-14 15:36:43
How are we going to do that?
AbhiWwis 2020-05-14 15:36:55
now we look at all the scenarios
AbhiWwis 2020-05-14 15:36:55
cant we just count
MR_67 2020-05-14 15:36:55
list out the possibilities
rrusczyk 2020-05-14 15:37:09
That will work. In fact, that's how I solved it when I did the problem.
rrusczyk 2020-05-14 15:37:16
But Harvey told me that wasn't very sporting.
rrusczyk 2020-05-14 15:37:23
He told me there's a cooler way to finish, but he wouldn't show it to me. Let's see if we can find it together.
sdattilo2002 2020-05-14 15:37:45
find the heights that satisfy triangle inequality
v4913 2020-05-14 15:38:09
The Triangle inequality maybe?
harmonyguan 2020-05-14 15:38:09
triangle inequality
cindyzou 2020-05-14 15:38:10
triangle inequality
matharpmath 2020-05-14 15:38:17
Triangle inequality
lrjr24 2020-05-14 15:38:17
triangle inequality
rrusczyk 2020-05-14 15:38:19
Oh yeah, we were talking about the Triangle Inequality a while back.
rrusczyk 2020-05-14 15:38:43
The Triangle Inequality tells us that we must have \begin{align*} a &< b+c\\ b &< a+c\\ c &< a+b\end{align*}in order for $a,b,c$ to be sides of a triangle.
doglover07 2020-05-14 15:39:22
we also know that a+b+c=h
christopherp 2020-05-14 15:39:22
a+b+c=h
jacobwu 2020-05-14 15:39:22
we alredady know that $a+b+c=h$
Chesssaga 2020-05-14 15:39:32
But it's easier to do with heights right?
rrusczyk 2020-05-14 15:40:08
Good idea -- we need to relate this to that interesting $a+b+c=h$ thing we just learned. What does this do to our Triangle Inequalities?
Mathworker2705 2020-05-14 15:40:51
We can substitute!!!
mmjguitar 2020-05-14 15:40:51
h-a=b+c
Jomo 2020-05-14 15:40:51
h-a>a
nikenissan 2020-05-14 15:41:00
2a < h, 2b < h, and 2c < h
rrusczyk 2020-05-14 15:41:10
We can replace $b+c$ with $h-a$, and similarly in the other two inequalities.
rrusczyk 2020-05-14 15:41:11
This gives \begin{align*} a &< h-a\\ b &< h-b\\ c &< h-c\end{align*}
rrusczyk 2020-05-14 15:41:14
And this becomes $2a < h$ and $2b < h$ and $2c < h$.
rrusczyk 2020-05-14 15:41:18
So for $a,b,c$ to be the sides of a triangle, they must all be less than $\dfrac{h}{2}$.
DaBobWhoLikeMath1234 2020-05-14 15:41:35
We've limited the possibilities!
RisingRook 2020-05-14 15:41:35
that narrows it down a lot
rrusczyk 2020-05-14 15:41:47
Yes we have!
rrusczyk 2020-05-14 15:41:57
What are the possibilities that work here?
ISolveLikeAGirl 2020-05-14 15:42:00
so then count the points that work
rrusczyk 2020-05-14 15:42:15
Yes, let's count 'em. Where are those points that work?
donguri 2020-05-14 15:42:50
so its in the middle quarter of the triangle
Jomo 2020-05-14 15:42:50
We can draw boundaries!
Jomo 2020-05-14 15:42:50
It can't be above the halfline in each direction!!
CHIPPER33 2020-05-14 15:42:50
the inner triangle
hi13 2020-05-14 15:42:50
draw a triangle to box in the possibilities
donguri 2020-05-14 15:42:50
the middle quarter of the triangle
ThatRichDeng 2020-05-14 15:42:50
if you draw all the midlines of the triangle the points in the middle region
rrusczyk 2020-05-14 15:42:54
We draw the three segments connecting the midpoints of the sides:
rrusczyk 2020-05-14 15:42:55
KevinW12345 2020-05-14 15:43:18
Somewhere around the center
mmjguitar 2020-05-14 15:43:18
around the center
fasterthanlight 2020-05-14 15:43:18
In the center
v4913 2020-05-14 15:43:18
Only in the very middle
Maths4J 2020-05-14 15:43:18
Divide it into 4 congruent equilateral triangles and the center one works (inside it)
Chesssaga 2020-05-14 15:43:32
Medial triangle!
rrusczyk 2020-05-14 15:43:37
Any point inside this smaller triangle will be less than $\dfrac{h}{2}$ from every side of the big triangle.
rrusczyk 2020-05-14 15:43:51
(And yes, we call that little triangle in the middle the "Medial triangle").
rrusczyk 2020-05-14 15:43:54
I think we found Harvey's solution!
KLBBC 2020-05-14 15:44:12
SO all the points in that region, which is 15??
OlympusHero 2020-05-14 15:44:12
5+4+3+2+1=15
T0-1h3MathMug360 2020-05-14 15:44:12
so its... 1+2+3+4+5=15
MathWizard10 2020-05-14 15:44:12
that is 15 points
RedFireTruck 2020-05-14 15:44:12
15
Wizkid8402 2020-05-14 15:44:12
$15$
sixoneeight 2020-05-14 15:44:12
there are 15 "good" points
universeking 2020-05-14 15:44:12
15 points work!
CT17 2020-05-14 15:44:12
So there are $\frac{5\cdot 6}{2}=15$ points that work!
rrusczyk 2020-05-14 15:44:25
Counting up the rows, we see that $5+4+3+2+1 = 15$ lattice points are inside that little triangle.
rrusczyk 2020-05-14 15:44:41
rponda 2020-05-14 15:44:56
1/3
MathematicianFlippityboii 2020-05-14 15:45:00
15/45=1/3!
BakedPotato66 2020-05-14 15:45:00
15/45= 1/3
yogurt12345 2020-05-14 15:45:00
15/45 is 1/3
universeking 2020-05-14 15:45:00
probability is 15/45=1/3
yoonsunam 2020-05-14 15:45:00
15/45=1/3
Shravi 2020-05-14 15:45:00
15/45=1/3
btc433 2020-05-14 15:45:00
The probability is $\frac{15}{45}=\frac13$
bobthegod78 2020-05-14 15:45:00
possible /total 15/45 =1/3 !!!
15/45 = 1/3
rrusczyk 2020-05-14 15:45:07
The probability is $\dfrac{\text{points that work}}{\text{total points}} = \dfrac{15}{45} = \boxed{\dfrac13}.$
rrusczyk 2020-05-14 15:45:19
Phew. One down.
mathgirl199 2020-05-14 15:45:27
Wow, that was actually pretty quick.
rrusczyk 2020-05-14 15:45:30
rrusczyk 2020-05-14 15:45:42
All right, a reminder: there are over 1300 of you here!
rrusczyk 2020-05-14 15:46:13
So, we aren't going to be able to react to all of your questions or comments today
rrusczyk 2020-05-14 15:46:22
Our usual classes are much, much smaller!!!
jabuticaba149 2020-05-14 15:46:33
That helped a lot thank you!
math12345678 2020-05-14 15:46:33
wasn't as hard as I think
jason543 2020-05-14 15:46:33
yay!!!!
Anniboy 2020-05-14 15:46:33
woohoo! Nice job! We solved it!
Duckyduckcheese 2020-05-14 15:46:33
YAY
mohanty 2020-05-14 15:46:37
Next Question!
CosmoMonkhouse 2020-05-14 15:46:37
many to go
DarthMaul 2020-05-14 15:46:39
whats next
rrusczyk 2020-05-14 15:46:50
That's one of my favorite lines...
rrusczyk 2020-05-14 15:46:51
Sprint #26: A circle is tangent to the positive $x$-axis at $x = 3.$ It passes through the distinct points $(6, 6)$ and $(p, p).$ What is the value of $p?$ Express your answer as a common fraction.
rrusczyk 2020-05-14 15:47:01
PerfectMath 2020-05-14 15:47:21
make a graph??
rondoallaturca 2020-05-14 15:47:21
graph what we can
doglover07 2020-05-14 15:47:21
draw a picture
coldcrazylogic 2020-05-14 15:47:21
draw?
flissyquokka17 2020-05-14 15:47:21
draw a diagram, right?
PoisonApple 2020-05-14 15:47:25
drawing it
rrusczyk 2020-05-14 15:47:38
I can't get anywhere without a picture. I'm not Harvey.
rrusczyk 2020-05-14 15:47:40
Let's sketch a picture of what's going on.
rrusczyk 2020-05-14 15:47:42
greenmathcounts 2020-05-14 15:48:14
center is (3,y)
asbodke 2020-05-14 15:48:14
The center of the circle lies on x=3
dineshs 2020-05-14 15:48:14
How do you assume that p<3
rrusczyk 2020-05-14 15:49:23
Indeed the center will be on the vertical line through $x=3$. This gives us the answer to the question dineshs asks (which is very much a good, non-obvious question!)
ccu700037 2020-05-14 15:49:37
p must be less than 3 because it must fall on y=x, and a line can only cross a circle at 2 points
DZL1 2020-05-14 15:49:37
and p,p and 6,6 are both on the line y=x... oh
rrusczyk 2020-05-14 15:50:02
Exactly -- $x=y$ hits our circle at two points, one on either side of the line $x=3$.
rrusczyk 2020-05-14 15:50:17
So, now we have:
rrusczyk 2020-05-14 15:50:23
rrusczyk 2020-05-14 15:51:48
Yikes -- I skipped a bit of step there. Why is the $y$-coordinate of the center of the circle equal to our radius?
rrusczyk 2020-05-14 15:51:56
(Which I've clearly called $r$)
AwesomeLife_Math 2020-05-14 15:52:47
The distance from (3,r) to (6,6) must be equal to r.
nihao4112 2020-05-14 15:52:47
Because of tangency
LANSH 2020-05-14 15:52:47
TANGENT
Puffer13 2020-05-14 15:52:47
it is tangent to the x-axis
mathdolphin123 2020-05-14 15:52:47
The circle is tangent
tigerzhang 2020-05-14 15:52:49
it is tangent to the x-axis
BakedPotato66 2020-05-14 15:52:51
it's tangent to the x axis
rrusczyk 2020-05-14 15:52:52
Ah right, the circle is tangent to the $x$-axis, so the segment from the center to $(3,0)$ is a radius.
rrusczyk 2020-05-14 15:53:08
What's next?
Zoobat 2020-05-14 15:53:18
so we use the distance formula to find r?
rrusczyk 2020-05-14 15:53:20
How?
jacobwu 2020-05-14 15:53:33
we can come up with an equation?
rrusczyk 2020-05-14 15:53:40
Good plan. How do we get the equation?
BBC0506 2020-05-14 15:53:48
Use the distance formula to find the distance from (3,r) to (6,6)
superagh 2020-05-14 15:53:48
distance to (6, 6) is equal to r
Maths4J 2020-05-14 15:53:48
Distance formula from (3, r) to (6, 6), then solve for r.
scinderella220 2020-05-14 15:53:51
r=the Distance from (3,r) to (6,6)
vsamc 2020-05-14 15:53:53
distance from (3,r) to (6,6) must be equal to r
rrusczyk 2020-05-14 15:54:00
The center must be $r$ away from $(6,6).$
rrusczyk 2020-05-14 15:54:05
What equation does this give us?
youcandomathnow 2020-05-14 15:54:29
(6-3)^2 + (6-r)^2 = r^2
asbodke 2020-05-14 15:54:29
r^2 = (6-r)^2 + 9
Jomo 2020-05-14 15:54:29
$r^2=3^2+(6-r)^2$
$r^2=3^2+(6-r)^2$
Speedstorm 2020-05-14 15:54:39
$(6-3)^2+(6-r)^2=r^2$
christopherp 2020-05-14 15:54:43
(6-3)^2 + (6-r)^2 = r^2
rrusczyk 2020-05-14 15:54:47
We must have $(6-3)^2 + (6-r)^2 = r^2.$
rrusczyk 2020-05-14 15:54:58
Time for some algebra.
ZippityA 2020-05-14 15:55:04
Expand.
jacobwu 2020-05-14 15:55:09
expand it
Piano_Man123 2020-05-14 15:55:21
Expand
ttmmrryy 2020-05-14 15:55:22
expand
rrusczyk 2020-05-14 15:55:25
Keep going!
Jomo 2020-05-14 15:55:43
$r^2=r^2-12r+45$
vsurya 2020-05-14 15:55:51
9+36-12r+r^2=r^2
menlo 2020-05-14 15:55:52
9+36 - 12r+r^2=r^2
winterrain01 2020-05-14 15:55:54
9+36-12r+r^2=r^2
rrusczyk 2020-05-14 15:55:59
Simplifying the left-hand side gives $r^2 - 12r + 45 = r^2.$
xMidnightFirex 2020-05-14 15:56:22
$r = \frac{15}4$
bobthegod78 2020-05-14 15:56:22
12r=45
Scipow 2020-05-14 15:56:22
r = 15/4
hi13 2020-05-14 15:56:22
12r=45
bobthegod78 2020-05-14 15:56:22
12r=45
KLBBC 2020-05-14 15:56:22
r=15/4
yizhou76 2020-05-14 15:56:22
r=15/4
Raiders26 2020-05-14 15:56:26
15/4=r
Mathqueen2018 2020-05-14 15:56:26
r = 15/4
rrusczyk 2020-05-14 15:56:30
The $r^2$'s cancel, and what we're left with gives $r= \dfrac{45}{12} = \dfrac{15}{4}.$
rrusczyk 2020-05-14 15:56:32
rrusczyk 2020-05-14 15:56:37
Now what?
Bryguy 2020-05-14 15:57:06
use distance formula again
Chesssaga 2020-05-14 15:57:06
Now use distance formula to calculate p
walrus987 2020-05-14 15:57:06
distance formula with p?
mathgirl199 2020-05-14 15:57:09
Now use distance formula for (3,15/4) and (p,p)
rrusczyk 2020-05-14 15:57:11
Often when a strategy works once in a problem, it will work again.
AwesomeLife_Math 2020-05-14 15:57:18
Distance formula on (p,p).
ThatRichDeng 2020-05-14 15:57:18
distance formula again
math12345678 2020-05-14 15:57:20
do the same thing to (p,p)
rrusczyk 2020-05-14 15:57:25
The distance from $(p,p)$ to the center must be $r$ too!
rrusczyk 2020-05-14 15:57:26
And we can write an equation for that, too.
rrusczyk 2020-05-14 15:57:28
$(p-3)^2+\left(p-\frac{15}{4}\right)^2 = \left(\frac{15}{4}\right)^2.$
rrusczyk 2020-05-14 15:57:46
What might you do first on this to make it easier to deal with?
Alculator11 2020-05-14 15:58:11
Multiply by 16
MathJams 2020-05-14 15:58:11
multiply both sides by 16?
Bloppity 2020-05-14 15:58:11
Get rid of fractions?
Kevin06 2020-05-14 15:58:11
Multiply by 16?
winterrain01 2020-05-14 15:58:11
get rid of the fraction?
TheEpicCarrot7 2020-05-14 15:58:11
Multiply by 16!
Puffer13 2020-05-14 15:58:11
multiply bot hsides by 16
rrusczyk 2020-05-14 15:58:16
I'd start by multiplying both sides by $16$, to get rid of the fractions.
rrusczyk 2020-05-14 15:58:17
We get $16(p-3)^2 + (4p - 15)^2 = 225.$
rrusczyk 2020-05-14 15:58:26
The rest is algebra.
TMSmathcounts737 2020-05-14 15:58:42
Now expand.
Cozzmo 2020-05-14 15:58:42
expand that
cj13609517288 2020-05-14 15:58:42
expand
bobthefam 2020-05-14 15:58:42
Expand!
ElNoraa 2020-05-14 15:58:42
expand?
G.G.Otto 2020-05-14 15:58:42
now expand
mathgirl199 2020-05-14 15:58:42
Now we can expand!
Mathematician1010 2020-05-14 15:58:42
now expand and simplify
universeking 2020-05-14 15:58:42
expand
Asterlan 2020-05-14 15:58:42
Then expand
rrusczyk 2020-05-14 15:58:48
Now, we can expand the left side and simplify, to get $32p^2 - 216p +144=0.$
Zoobat 2020-05-14 15:58:56
4p^2 - 27p + 18 = 0
rrusczyk 2020-05-14 15:59:24
Dividing both sides by $8$ gives $4p^2 - 27p +18=0.$
asbodke 2020-05-14 15:59:33
(p-6)(4p-3) = 0
huela 2020-05-14 15:59:53
(4p-3)(p-6)=0
rrusczyk 2020-05-14 15:59:54
Factoring gives $(4p-3)(p-6)=0.$
rrusczyk 2020-05-14 16:00:00
Uh-oh. Two solutions.
doglover07 2020-05-14 16:00:26
and p is not 6
Chesssaga 2020-05-14 16:00:26
p = 3/4 since we proved it is not greater than 3
Bryguy 2020-05-14 16:00:26
p cannot equal 6
winterrain01 2020-05-14 16:00:27
p must be the smaller solution
rrusczyk 2020-05-14 16:00:59
Ah, that's right. We want the point that's not $(6,6)$. So $p$ is...
PV123 2020-05-14 16:01:19
3/4
mark888 2020-05-14 16:01:19
3/4!
hi13 2020-05-14 16:01:19
3/4
alphaKITTY37 2020-05-14 16:01:19
3/ 4
azhang1202 2020-05-14 16:01:19
3/4
RJ5303707 2020-05-14 16:01:19
$3/4$
BananaWatch5223 2020-05-14 16:01:19
3/4
ChessSavage 2020-05-14 16:01:19
3/4
colorpencilz 2020-05-14 16:01:19
3/4
MathWizard10 2020-05-14 16:01:19
3/4
bonami2018 2020-05-14 16:01:19
3/4
3/4
christopherp 2020-05-14 16:01:19
3/4
arisettyjr 2020-05-14 16:01:19
3/4
WAWATHEGOAT 2020-05-14 16:01:19
3/4
iam_awesome 2020-05-14 16:01:23
$$3/4$$
joseph2718 2020-05-14 16:01:23
3/4
robot317 2020-05-14 16:01:23
p is 3/4
rrusczyk 2020-05-14 16:01:28
So we take the solution $p = \boxed{\dfrac34}.$
TheEpicCarrot7 2020-05-14 16:01:50
Yay we solved it!
scinderella220 2020-05-14 16:01:50
YAY!
Mistyketchum28 2020-05-14 16:01:50
Yay!
tacos4life123 2020-05-14 16:01:50
yeah
Terribleteeth 2020-05-14 16:01:50
yay! we did it
CHIPPER33 2020-05-14 16:01:50
YAY! NEXT
Raiders26 2020-05-14 16:01:50
Yay!!!
sahila 2020-05-14 16:01:50
yay
MathWizard38025 2020-05-14 16:01:50
thats a good solution
Mistyketchum28 2020-05-14 16:01:54
That was a good one!
SandyBeach11 2020-05-14 16:01:54
YAY WE DID IT
sahanapotu 2020-05-14 16:01:57
now it amkes sense
rrusczyk 2020-05-14 16:02:10
Great!
punkinpiday 2020-05-14 16:02:14
Two down, four to go!
rrusczyk 2020-05-14 16:02:25
Here's the next one:
rrusczyk 2020-05-14 16:02:25
Sprint #27: How many of the first $2019$ positive integers have no odd single-digit prime factors?
v4913 2020-05-14 16:02:47
THIS WAS AN AWESOME PROBLEM XD
rrusczyk 2020-05-14 16:02:52
Equinox8 2020-05-14 16:03:18
complementary counting?
srim1027 2020-05-14 16:03:18
use complementary counting?
ryanfu2008 2020-05-14 16:03:18
complementary counting???
awsomek 2020-05-14 16:03:18
complementary count?
rrusczyk 2020-05-14 16:03:32
For those of you who don't know this term:
MR_67 2020-05-14 16:03:34
do we count the "not"
rrusczyk 2020-05-14 16:03:50
"Complementary counting" is a fancy way to say "count the thing you don't want".
rrusczyk 2020-05-14 16:04:04
What's a clue in this problem that we might want to count the things we don't want?
harmonyguan 2020-05-14 16:04:33
"no"
RisingRook 2020-05-14 16:04:33
the word no
panda2018 2020-05-14 16:04:33
'no'
Monkey287 2020-05-14 16:04:33
have NO
green_alligator 2020-05-14 16:04:33
it says "have no"
HappySpring 2020-05-14 16:04:33
the "no"
doglover07 2020-05-14 16:04:33
"no"
DottedCaculator 2020-05-14 16:04:33
NO
AlphaBetaGammaOmega 2020-05-14 16:04:33
the word "no"
aops_band 2020-05-14 16:04:40
no
iam_awesome 2020-05-14 16:04:40
the word "no"
razormouth 2020-05-14 16:04:40
the word "no"
Math5K 2020-05-14 16:04:40
no
MyNameIsJeffZF 2020-05-14 16:04:40
NO odd ...
superagh 2020-05-14 16:04:40
"no"
rrusczyk 2020-05-14 16:04:58
"No" and "not" are clues. There's also just the number of cases to think about.
rrusczyk 2020-05-14 16:05:05
What are the factors we have to avoid?
asbodke 2020-05-14 16:05:18
3,5,7
jupiter314 2020-05-14 16:05:18
3,5 and 7
fasterthanlight 2020-05-14 16:05:18
3, 5, and 7
NHatAOPS 2020-05-14 16:05:18
3 5 and 7
sv739 2020-05-14 16:05:18
3,5,7
Leonard_my_dude 2020-05-14 16:05:18
3,5,7
n3v3rmor3 2020-05-14 16:05:18
3, 5, 7
Jerry_Guo 2020-05-14 16:05:18
3,5,7
AMC_Kid 2020-05-14 16:05:18
3, 5, 7
math_piggy 2020-05-14 16:05:18
3, 5, and 7
os31415 2020-05-14 16:05:18
3, 5, and 7
yayatheduck 2020-05-14 16:05:18
3, 5, 7
snorlax186 2020-05-14 16:05:18
3,5, and 7
SkywalkerAUV 2020-05-14 16:05:18
3, 5, and 7
$3, 5, 7$
tenebrine 2020-05-14 16:05:21
3, 5, 7
Mathqueen2018 2020-05-14 16:05:21
3, 5, 7
Speedstorm 2020-05-14 16:05:22
3,5 and 7
maxxie2019 2020-05-14 16:05:24
3, 5, 7
C0atimundi 2020-05-14 16:05:24
3, 5, 7
Minibangs 2020-05-14 16:05:24
3, 5, and 7
rrusczyk 2020-05-14 16:05:28
We need to avoid the odd single-digit primes, which are $3$, $5$, and $7.$
rrusczyk 2020-05-14 16:05:39
That leave us with all the numbers that only have factors $2, 11, 13, 17,$ etc., but that's a lot of primes to keep track of! I don't even know how many there are!
rrusczyk 2020-05-14 16:06:02
So, the stuff we want to count covers a TON of cases. The stuff we don't want just covers 3 primes.
sixoneeight 2020-05-14 16:06:11
Subtract the cases you don't want from the total
rrusczyk 2020-05-14 16:06:27
That's the plan. Let's count how many of the first $2019$ positive integers have $3, 5,$ or $7$ as a factor. This counts the numbers we don't want.
rrusczyk 2020-05-14 16:06:33
Then we'll subtract this count from $2019$ to get our answer: the count of the numbers we do want.
rrusczyk 2020-05-14 16:06:35
How many of them have $3$ as a factor?
smartatmath 2020-05-14 16:06:57
673
The_Better_Samuel 2020-05-14 16:06:57
673
aie8920 2020-05-14 16:06:57
673
Ninja11 2020-05-14 16:06:57
673
Lux1 2020-05-14 16:06:57
there are 673
Jomo 2020-05-14 16:06:57
673
nye 2020-05-14 16:06:57
673
HappySpring 2020-05-14 16:06:57
673
AwesomeLife_Math 2020-05-14 16:06:57
673.
bot101 2020-05-14 16:06:57
673
lihao_david 2020-05-14 16:06:57
673
dhaops 2020-05-14 16:06:57
673
Awesome_360 2020-05-14 16:06:57
673
krishgarg 2020-05-14 16:06:57
673
Meyzeek_Saveer 2020-05-14 16:06:57
673
Happytwin 2020-05-14 16:06:57
673
Quaoar 2020-05-14 16:06:57
There are 673 multiples of 3
IAmTheHazard 2020-05-14 16:07:07
Floor(2019/3)=673
rrusczyk 2020-05-14 16:07:10
We compute that $2019 \div 3 = 673.$
rrusczyk 2020-05-14 16:07:12
So there are $673$ numbers in our set that are a multiple of $3.$
rrusczyk 2020-05-14 16:07:13
How many of them have $5$ as a factor?
KOYO 2020-05-14 16:07:43
403
bluemathcounts 2020-05-14 16:07:43
403
justinzhang404 2020-05-14 16:07:43
403
SparklyFlowers 2020-05-14 16:07:43
403
nose 2020-05-14 16:07:43
403
robot317 2020-05-14 16:07:43
403
krithikrokcs 2020-05-14 16:07:43
403
cat_maniac_8 2020-05-14 16:07:43
403
melodyzhao 2020-05-14 16:07:43
403
twinbrian 2020-05-14 16:07:43
$403$
scoutskylar 2020-05-14 16:07:43
403
rayfish 2020-05-14 16:07:43
$\lfloor \frac{2019}{5}\rfloor = 403$
Pokemon2 2020-05-14 16:07:43
floor(2019/5)=403
tejasgandi 2020-05-14 16:07:43
403
palindrome868 2020-05-14 16:07:43
403
dineshs 2020-05-14 16:07:43
2019/5 = 403.8, so 403
rrusczyk 2020-05-14 16:07:48
We compute that $2019 \div 5 = 403\frac45.$
rrusczyk 2020-05-14 16:07:57
How many of them have $7$ as a factor?
MathMaster029 2020-05-14 16:08:18
288
yizhou76 2020-05-14 16:08:18
288
hungrypig 2020-05-14 16:08:18
288
TheEpicCarrot7 2020-05-14 16:08:18
288
ElNoraa 2020-05-14 16:08:18
288
twinklebrownie 2020-05-14 16:08:18
288
Math5K 2020-05-14 16:08:18
288
PureSwag 2020-05-14 16:08:18
$\lfloor \frac{2019}{7} \rfloor = 288$
WAWATHEGOAT 2020-05-14 16:08:18
288
bobthefam 2020-05-14 16:08:18
288
douglasubella 2020-05-14 16:08:18
288
hxz913 2020-05-14 16:08:18
288
NonOxyCrustacean 2020-05-14 16:08:18
288
Awesome_360 2020-05-14 16:08:18
288
gordonhero 2020-05-14 16:08:18
288
enkitty 2020-05-14 16:08:18
Floor(2019/7)=288
rrusczyk 2020-05-14 16:08:21
We compute that $2019 \div 7 = 288\frac37.$
rrusczyk 2020-05-14 16:08:22
So there are $288$ numbers in our set that are a multiple of $7.$
rrusczyk 2020-05-14 16:08:27
It looks like this gives us $673+403+288 = 1364$ numbers total that we don't want. Is that right?
pog 2020-05-14 16:08:40
It's just a venn diagram question really
rrusczyk 2020-05-14 16:09:06
Oh, interesting comment --- Venn diagrams have...
dws1188 2020-05-14 16:09:14
overlap
RedFireTruck 2020-05-14 16:09:14
OVERLAPS
rrusczyk 2020-05-14 16:09:17
Overlaps.
joseph2718 2020-05-14 16:09:32
No, we're overcounting
rayfish 2020-05-14 16:09:32
No, we are overcounting
NHatAOPS 2020-05-14 16:09:32
WE OVERCOUNTED!
Raiders26 2020-05-14 16:09:32
overcounting
MathJams 2020-05-14 16:09:32
no we overcounted
vsamc 2020-05-14 16:09:32
no but we are overcounting
Juno 2020-05-14 16:09:32
WE OVERCOUNTED
MathyAOP 2020-05-14 16:09:32
Overcounting
knoxboy 2020-05-14 16:09:32
some numbers overlap
aopscasanjose 2020-05-14 16:09:32
no you overcounted
rrusczyk 2020-05-14 16:09:43
We overcounted! We counted some numbers twice.
Alculator11 2020-05-14 16:09:48
rrusczyk 2020-05-14 16:09:57
Yeah, the multiples of 21. And what else?
not_kevin_888 2020-05-14 16:10:23
15 and 35
bobthegod78 2020-05-14 16:10:23
35, and 15
pranavaggarwal 2020-05-14 16:10:23
35
ZippityA 2020-05-14 16:10:23
and 35
coldcrazylogic 2020-05-14 16:10:23
15, 35
aaronw 2020-05-14 16:10:23
15 and 35
Chesssaga 2020-05-14 16:10:23
Multiples of 15
christopherp 2020-05-14 16:10:23
15, 35
Doodles250 2020-05-14 16:10:23
and 35
Dumdum990 2020-05-14 16:10:23
multiples of 15
tumbleweed 2020-05-14 16:10:23
15 and 35
GrizzyProblemSolver79c 2020-05-14 16:10:23
multiples of 15 and 35
rrusczyk 2020-05-14 16:10:32
Yeah, multiples of 15 and 35 are problems, too.
rrusczyk 2020-05-14 16:10:55
Let's take a look at what happened with just one of these: $15.$ It has a $3$ and a $5$ in its prime factorization, so it was included twice in our total, once among the multiples of $3$ and once among the multiples of $5.$
rrusczyk 2020-05-14 16:10:58
The same is true of every multiple of $15.$
rrusczyk 2020-05-14 16:10:59
And of every multiple of $3\cdot 7=21,$ and of every multiple of $5\cdot 7 = 35.$
rrusczyk 2020-05-14 16:11:10
So, what do we do to fix this?
MLiang2018 2020-05-14 16:11:41
subtract
dineshs 2020-05-14 16:11:41
subtract
azhang1202 2020-05-14 16:11:41
We subtract them from our total
Subtract
obi-wan-the 2020-05-14 16:11:41
SUBTRACT THEM
AwesomeLife_Math 2020-05-14 16:11:41
Subtract them from our total.
NerdyDude 2020-05-14 16:11:41
subtract them
RJ5303707 2020-05-14 16:11:41
Subtract these cases
Speedstorm 2020-05-14 16:11:41
subtract each of them once
robot317 2020-05-14 16:11:41
subtract them
doglover07 2020-05-14 16:11:41
subtract those multiples
Zoobat 2020-05-14 16:11:41
subtract out those extra ones
OlympusHero 2020-05-14 16:11:41
Subtract amount of multiples of 15,21 and 35
capitanhanbo 2020-05-14 16:11:41
subtract the overlappers
rrusczyk 2020-05-14 16:12:02
We've counted each of these multiples twice in our total.
rrusczyk 2020-05-14 16:12:03
So we need to subtract the total number of these multiples once, to correct for this double counting.
rrusczyk 2020-05-14 16:12:05
How many multiples of $15$ did we double-count?
aops-as 2020-05-14 16:12:40
134
happyhari 2020-05-14 16:12:40
134
Jalenluorion 2020-05-14 16:12:40
134
sdattilo2002 2020-05-14 16:12:40
134
millburn2006 2020-05-14 16:12:40
134
MathBluebird 2020-05-14 16:12:40
134
math_piggy 2020-05-14 16:12:40
$134$
pinkmathcounts 2020-05-14 16:12:40
134
sd-pear 2020-05-14 16:12:40
134
Elaine_Wang 2020-05-14 16:12:40
134
messimagic 2020-05-14 16:12:40
134
Math-Marvel 2020-05-14 16:12:40
floor(2019/15)
FireDragon1719 2020-05-14 16:12:40
$floor2019/15$
rrusczyk 2020-05-14 16:12:43
We have $2019 \div 15 = 134\frac{9}{15},$ so there are $134$ multiples of $15$ that we double-counted in our original total.
rrusczyk 2020-05-14 16:12:44
How about multiples of $21?$
melonlord 2020-05-14 16:13:11
96
lucky0318 2020-05-14 16:13:11
96
scinderella220 2020-05-14 16:13:11
96
Math4Life7 2020-05-14 16:13:11
96
bryding21 2020-05-14 16:13:11
96
AWaltz 2020-05-14 16:13:11
96
Quentissential 2020-05-14 16:13:11
96
Eth007 2020-05-14 16:13:11
96
agentmath 2020-05-14 16:13:11
96
KevinW12345 2020-05-14 16:13:11
96
DerpyTaterTot 2020-05-14 16:13:11
96
PShucks 2020-05-14 16:13:11
96
Styxxx 2020-05-14 16:13:11
96
ellnoo 2020-05-14 16:13:11
96
sarahAops2020 2020-05-14 16:13:11
96
kzhang88 2020-05-14 16:13:11
96
math12345678 2020-05-14 16:13:11
96
PureSwag 2020-05-14 16:13:11
$\lfloor \frac{2019}{21} \rfloor = 96$
razormouth 2020-05-14 16:13:11
96
96
rrusczyk 2020-05-14 16:13:14
We have $2019 \div 21 = 96\frac{3}{21},$ so there are $96$ multiples of $21$ that we double-counted in our original total.
rrusczyk 2020-05-14 16:13:14
How about multiples of $35?$
huela 2020-05-14 16:13:34
57
matt1010 2020-05-14 16:13:34
57 for 35
NerdyDude 2020-05-14 16:13:34
57
Mandakini 2020-05-14 16:13:34
57
bobthefam 2020-05-14 16:13:34
floor(2019/35)=57
ZippityA 2020-05-14 16:13:34
57
mathlete5451006 2020-05-14 16:13:34
57
Eng123 2020-05-14 16:13:34
57
bedwinprusik578 2020-05-14 16:13:34
57
smartatmath 2020-05-14 16:13:34
57]
Scipow 2020-05-14 16:13:34
57
jjw4106 2020-05-14 16:13:34
57
srm2005 2020-05-14 16:13:34
57
s2010n 2020-05-14 16:13:34
57
mathematics10237 2020-05-14 16:13:34
57
Knightofaops 2020-05-14 16:13:34
57
rrusczyk 2020-05-14 16:13:37
We have $2019 \div 35 = 57\frac{24}{35},$ so there are $57$ multiples of $35$ that we double-counted in our original total.
rrusczyk 2020-05-14 16:13:39
So, we have to take away $134+96+57 = 287$ from our earlier count, which gives us a new count of $1364 - 287 = 1077.$
Doudou_Chen 2020-05-14 16:13:54
how about multiples of $105 = 3\cdot5\cdot7$?
m_goli 2020-05-14 16:13:54
MathWiz20 2020-05-14 16:13:54
NHatAOPS 2020-05-14 16:13:57
CESAJ 2020-05-14 16:14:01
but what about multiples of 105?
rrusczyk 2020-05-14 16:14:14
What's so important about $105$?
MathJams 2020-05-14 16:14:37
multiple of 3,5 and 7
Math5K 2020-05-14 16:14:37
divisible by 3,5,7
srim1027 2020-05-14 16:14:37
it is the lcm of 3 5 7
MyNameIsJeffZF 2020-05-14 16:14:37
3*5*7
mathcounts_genius 2020-05-14 16:14:37
3*5*7
mmjguitar 2020-05-14 16:14:37
its 5*3*7
yayatheduck 2020-05-14 16:14:37
it's 3*5*7
anshultheking1 2020-05-14 16:14:37
3x5x7
Shmileyface 2020-05-14 16:14:37
multiple of 3, 5, and 7
olive0827 2020-05-14 16:14:37
3*5*7
enkitty 2020-05-14 16:14:37
It's a multiple of all three.
huangtx2018 2020-05-14 16:14:39
it's the LCM of 3, 5, and 7
DonutHole 2020-05-14 16:14:39
It is a multiple of 3, 5, and 7
rrusczyk 2020-05-14 16:14:44
It is a multiple of all of $3, 5,$ and $7.$
rrusczyk 2020-05-14 16:14:53
Hmmmm... How many times have we counted it so far?
huela 2020-05-14 16:15:18
0
scoutskylar 2020-05-14 16:15:18
0
RJ5303707 2020-05-14 16:15:18
None
zhuqingzhang 2020-05-14 16:15:18
0
Math5K 2020-05-14 16:15:18
0
rayfish 2020-05-14 16:15:18
0 times!!!
Pokemon2 2020-05-14 16:15:18
0 times
missionsqhc 2020-05-14 16:15:18
0
rrusczyk 2020-05-14 16:15:38
None? But I'm pretty sure we counted it with the multiples of 3. And the multiples of 5. And the multiples of 7.
scoutskylar 2020-05-14 16:16:04
We counted them three times, but un-counted them three times.
arirah9 2020-05-14 16:16:04
3 times, and then we subtracted it 3 times, so 0
DottedCaculator 2020-05-14 16:16:04
but then we subtracted them
odinthunder 2020-05-14 16:16:04
But then we subtracted 3 times
green_alligator 2020-05-14 16:16:04
but we subtracted it 3 times as well
lilyba 2020-05-14 16:16:06
we then subtracted it three times
user0003 2020-05-14 16:16:08
subtracted three times
rrusczyk 2020-05-14 16:16:38
Oh, yeah. We took it back away with the multiples of 15. And the multiples of 21. And the multiples of 35.
SaltyCracker 2020-05-14 16:16:48
it overlaps with the other overlappers
rrusczyk 2020-05-14 16:16:56
I could not possibly have said it better myself.
rrusczyk 2020-05-14 16:17:05
It got counted three times in our original count (once for each prime), and then subtracted three times in our correction (once for each product of two primes).
rrusczyk 2020-05-14 16:17:07
That means that in our total of $1077,$ the number $105$ (and all multiples of $105$) aren't counted at all!
Mistyketchum28 2020-05-14 16:17:20
We have to add it back once!
DonutHole 2020-05-14 16:17:20
So we have to add it back
anishcool11 2020-05-14 16:17:24
nikenissan 2020-05-14 16:17:26
We need to add back that number of 105 multiples
mark888 2020-05-14 16:17:34
oh no, gotta add it back
Chesssaga 2020-05-14 16:17:34
bluemathcounts 2020-05-14 16:17:36
poopypurplebob 2020-05-14 16:17:38
rrusczyk 2020-05-14 16:17:41
We have to add them back in.
fasterthanlight 2020-05-14 16:17:51
There are 19 of them
NHatAOPS 2020-05-14 16:17:51
there's 19 multiples
MLTC 2020-05-14 16:17:51
jupiter314 2020-05-14 16:17:51
CHIPPER33 2020-05-14 16:17:51
MathMaster029 2020-05-14 16:17:51
Maths4J 2020-05-14 16:17:51
There are 19 of them
enkitty 2020-05-14 16:17:51
There are 19 multiples of 105, so add it to 1077
Asterlan 2020-05-14 16:17:53
19 multiples of it to add back!
rrusczyk 2020-05-14 16:17:56
Since $2019 \div 105 = 19\frac{24}{105},$ we have $19$ multiples of $105$ less than $2019.$ (It's probably easier to just notice that $20 \cdot 105 = 2100$ is too big, and $19 \cdot 105 = 2100 - 105 = 1995$ is small enough.)
rrusczyk 2020-05-14 16:17:59
We add these back in to our running total and we have $1077+19 = 1096$ total numbers with at least one odd single-digit prime factor.
rrusczyk 2020-05-14 16:18:12
Phew, so do we write down 1096 and move on now?
HappySpring 2020-05-14 16:18:47
we have to subtract it from 2019
mathboykl2018 2020-05-14 16:18:47
subtract that from 2019
AwesomeLife_Math 2020-05-14 16:18:47
Subtract that from 2019 to get the answer.
ARSM2019 2020-05-14 16:18:47
2019-1096
Jomo 2020-05-14 16:18:47
wait! we are complementary counting!!!
bobthegod78 2020-05-14 16:18:47
2019-1096=923!!
NHatAOPS 2020-05-14 16:18:47
subtract that from 2019 to get the answer
joseph2718 2020-05-14 16:18:47
2019 - 1096 !!
Now do $2019-1096$
anc3 2020-05-14 16:18:47
Noooo subtract from total of 2019 because we used complementary counting
palindrome868 2020-05-14 16:18:47
2019 - 1096 = 923!!!
RedFireTruck 2020-05-14 16:18:48
2019-1096 = 923
rrusczyk 2020-05-14 16:18:56
Oh yeah! I'm glad y'all are on my team!
rrusczyk 2020-05-14 16:19:02
We've now accurately counted each multiple of $3, 5,$ or $7$ exactly once.
rrusczyk 2020-05-14 16:19:04
But remember these are the numbers we don't want!
rrusczyk 2020-05-14 16:19:05
We have to subtract this count from $2019$ to get our final answer.
PShucks 2020-05-14 16:19:15
akpi 2020-05-14 16:19:24
2019-1096=923
lordoftherings4926 2020-05-14 16:19:24
We write down 923
Speedstorm 2020-05-14 16:19:24
2019-1096=923
hunthunthunt 2020-05-14 16:19:24
$923$
dhaops 2020-05-14 16:19:29
923.
jonahc426 2020-05-14 16:19:29
923
Raiders26 2020-05-14 16:19:29
923
sixoneeight 2020-05-14 16:19:30
Dumdum990 2020-05-14 16:19:30
923 possible outcomes
robot317 2020-05-14 16:19:33
rrusczyk 2020-05-14 16:19:35
Therefore, $2019-1096 = \boxed{923}$ of the first $2019$ positive integers have no odd single-digit prime factors.
ZOP123 2020-05-14 16:19:44
YAY
Math4Life7 2020-05-14 16:19:44
yayyyyy
rrusczyk 2020-05-14 16:19:53
All right -- DPatrick's turn!
DPatrick 2020-05-14 16:19:59
Thanks -- hi again!
DPatrick 2020-05-14 16:20:07
Sprint #28: What is the value of $\sqrt{11{,}111{,}111\times 100{,}000{,}011+4}?$
DPatrick 2020-05-14 16:20:21
Remember: no calculators!
nathanqiu 2020-05-14 16:20:47
that looks terrifying!
kaciecheng 2020-05-14 16:20:50
DPatrick 2020-05-14 16:20:59
Yikes. I don't want to multiply out those big numbers, add $4,$ and try to take a square root.
DPatrick 2020-05-14 16:21:09
Any better ideas?
sixoneeight 2020-05-14 16:21:31
Try to find a pattern!
menlo 2020-05-14 16:21:42
100000011=100000000+11
DPatrick 2020-05-14 16:22:01
I like this thought. Are there easier ways we can write those big numbers?
Makemakika1 2020-05-14 16:22:15
iiRishabii 2020-05-14 16:22:31
Split 100000011 into (100000000 + 11)
Qiu10 2020-05-14 16:22:36
100000011 is 100000000+11
duansaops 2020-05-14 16:22:36
11111111 = (10^8-1)/9
DPatrick 2020-05-14 16:22:45
Aha!
DPatrick 2020-05-14 16:22:57
$11{,}111{,}111$ is one-ninth of $99{,}999{,}999,$ which is one less than $100{,}000{,}000 = 10^8.$
DPatrick 2020-05-14 16:23:16
So we can write $11{,}111{,}111 = \dfrac{10^8-1}{9}.$
DPatrick 2020-05-14 16:23:25
That seems like it'll be a lot easier to work with.
Jomo 2020-05-14 16:23:40
and $10^8+11$
sixoneeight 2020-05-14 16:23:40
so its 10^8+11
MR_67 2020-05-14 16:23:40
100000011 = 10^8+11, 11111111=(10^8-1)/9
DPatrick 2020-05-14 16:23:51
Good idea. Especially since we have $10^8$ in our first number, I think $10^8 + 11$ is a lot easier to work with for the second number.
DPatrick 2020-05-14 16:24:03
So now our expression is $$\sqrt{\dfrac{10^8-1}{9} \times (10^8+11) + 4}.$$
DPatrick 2020-05-14 16:24:12
This seems much simpler to work with!
DPatrick 2020-05-14 16:24:19
What now?
Alexm12 2020-05-14 16:24:47
9 is a square!!
ajax31 2020-05-14 16:24:47
9 is 3²
Scipow 2020-05-14 16:24:47
factor out the 1/9!
Ninja-Girl 2020-05-14 16:25:00
$9$ = $3^2$
enkitty 2020-05-14 16:25:03
Could we multiply by 9 so we don't have to deal with fractions
DPatrick 2020-05-14 16:25:19
Well, that $9$ is a tantalizing perfect square just sitting there.
DPatrick 2020-05-14 16:25:36
But it's hard to work with, seeing as it's sitting there by itself.
scoutskylar 2020-05-14 16:25:45
That's equal to $\frac{1}{3}\sqrt{\left(10^{8}-1\right)\cdot\left(10^{8}+11\right)+36}$
DottedCaculator 2020-05-14 16:25:57
4=36/9
Alculator11 2020-05-14 16:25:59
Multiply $4$ by $9/9$
Raiders26 2020-05-14 16:26:10
make denominator 9
DPatrick 2020-05-14 16:26:16
Good idea.
DPatrick 2020-05-14 16:26:26
Let's make it a single fraction!
DPatrick 2020-05-14 16:26:31
We get $\sqrt{\dfrac{(10^8-1)(10^8+11)+36}{9}}.$
DPatrick 2020-05-14 16:26:47
And now we can pull the $9$ outside if we want.
DPatrick 2020-05-14 16:27:08
So we have $\dfrac13\sqrt{(10^8-1)(10^8+11)+36}.$
DPatrick 2020-05-14 16:27:19
Now what?
asbodke 2020-05-14 16:27:28
Multiply it out
MathJams 2020-05-14 16:27:28
expand $(10^8-1)(10^8+11)+36$!
ThatRichDeng 2020-05-14 16:27:36
expand it
dolphin7 2020-05-14 16:27:36
expand
doglover07 2020-05-14 16:27:36
expand
user0003 2020-05-14 16:27:36
expand
agentmath 2020-05-14 16:27:36
expand
melonlord 2020-05-14 16:27:36
expand
KingRavi 2020-05-14 16:27:36
expand
DPatrick 2020-05-14 16:27:48
Sure, we can multiply the two binomials together.
DPatrick 2020-05-14 16:27:57
What do we get under the square root when we do that?
Speedstorm 2020-05-14 16:28:44
The root simplifies to $10^{16}+10^9+25$
AbhiWwis 2020-05-14 16:28:44
10^16 + 10(10^8) + 25
peace09 2020-05-14 16:28:44
$10^16+10^9+5^2$
sigma_notation 2020-05-14 16:28:44
multiply it out to (10^16+10^9+25)
palindrome868 2020-05-14 16:28:44
10^16 + 10^9 + 25
DPatrick 2020-05-14 16:29:01
We get $\dfrac13\sqrt{(10^8)^2 + 10(10^8) + 25}.$
MTHJJS 2020-05-14 16:29:33
the square root is = $(10^8 + 5)$
AbhiWwis 2020-05-14 16:29:33
(10^8 + 5)^2
Speedstorm 2020-05-14 16:29:33
$(10^8+5)^2$
duansaops 2020-05-14 16:29:33
$(10^8+5)^2$
Maths4J 2020-05-14 16:29:33
(10^8+5)^2
sdattilo2002 2020-05-14 16:29:33
nice!
srim1027 2020-05-14 16:29:33
square of a binomial
Quentissential 2020-05-14 16:29:33
(10^8 +5)^2
Dalar25 2020-05-14 16:29:33
perfect square!
nascar48 2020-05-14 16:29:33
perfect square
DPatrick 2020-05-14 16:29:39
Hey, that looks like a perfect square!
DPatrick 2020-05-14 16:29:55
Indeed, it's $\dfrac13\sqrt{\left(10^8 + 5\right)^2}.$
Imishkabob 2020-05-14 16:30:02
And then we take the perfect squares out
Math5K 2020-05-14 16:30:06
it's 1/3(10^8+5)
sixoneeight 2020-05-14 16:30:14
Wow! so we get (10^8+5)/3
vtlev 2020-05-14 16:30:16
(10^8+5)/3
DPatrick 2020-05-14 16:30:20
So our expression is just $\dfrac{10^8 + 5}{3}.$
DPatrick 2020-05-14 16:30:28
And how do we finish up?
superagh 2020-05-14 16:30:42
and then we divide
DottedCaculator 2020-05-14 16:30:42
=10000005/3=33333335
MR_67 2020-05-14 16:30:42
100000005/3 = 33333335
athik789 2020-05-14 16:30:42
100000005/3
Mistyketchum28 2020-05-14 16:30:45
100000005/3
Ninja11 2020-05-14 16:30:45
which is 100000005/3
GrizzyProblemSolver79c 2020-05-14 16:30:54
$100000005/3 = 33333335$
$\frac{100000005}{3}$
Lux1 2020-05-14 16:30:54
and then that comes out to 33,333,335
DPatrick 2020-05-14 16:31:00
We could certainly compute $100{,}000{,}005 \div 3,$ but let's use the same trick we used in the beginning.
DPatrick 2020-05-14 16:31:07
We know that $\dfrac{10^8-1}{3} = 33{,}333{,}333.$ How does that help?
b20081 2020-05-14 16:31:19
note that 9999999=333333*3
BakedPotato66 2020-05-14 16:31:23
$10^8-1$ is divisible by 3
b20081 2020-05-14 16:31:29
and then 9999999+6=10^8+5
GrizzyProblemSolver79c 2020-05-14 16:31:32
it might be easier to do $(99999999 + 6)/3$
PureSwag 2020-05-14 16:31:35
We can note that $99,999,999 \div 3 = 33,333,333$
MyNameIsJeffZF 2020-05-14 16:31:40
Wontoflonto 2020-05-14 16:31:40
huela 2020-05-14 16:31:40
+6/3
Puffer13 2020-05-14 16:31:40
DPatrick 2020-05-14 16:31:43
Exactly!
DPatrick 2020-05-14 16:31:51
We can write $$\dfrac{10^8+5}{3} = \dfrac{10^8-1+6}{3} = \dfrac{10^8-1}{3} + 2.$$
aaravdodhia 2020-05-14 16:32:10
Then $\frac{10^8 + 5}{3} = \frac{10^8 - 1}{3} + \frac{6}{3} = 33,333,333 + 2$.
C0atimundi 2020-05-14 16:32:10
Ninja11 2020-05-14 16:32:10
You can add 2 to 33333333
dolphin7 2020-05-14 16:32:15
$33,333,333+2=33,333,335‬$
CosmoMonkhouse 2020-05-14 16:32:20
All we have to do is add 2 to get 33333335
bobjoebilly 2020-05-14 16:32:20
33, 333, 335
DPatrick 2020-05-14 16:32:24
And then this is just $33{,}333{,}333 + 2 = \boxed{33{,}333{,}335}.$
DPatrick 2020-05-14 16:32:45
The moral of the story: big numbers are icky. Try to make them less icky.
bestzack66 2020-05-14 16:32:55
YAY we solved another problem!
Speedstorm 2020-05-14 16:32:55
that was a nice solution
ssr_07 2020-05-14 16:32:55
That was a good one.
MrAnishNagariya 2020-05-14 16:32:59
that looked so scary, not it seems so easy
DPatrick 2020-05-14 16:33:08
A lot of problems look scarier than they really are!
DPatrick 2020-05-14 16:33:23
I feel like the next problem was written especially for me!
DPatrick 2020-05-14 16:33:29
Sprint #29: David throws a dart at a triangular dartboard whose side lengths are $5, 5$ and $6,$ and the dart lands in a random location on the dartboard. What is the probability that the sum of the squares of the $3$ distances from the dart's location to the corners of the dartboard is less than $30?$ Express your answer as a common fraction in terms of $\pi.$
Eng123 2020-05-14 16:34:02
Geometric probability?
CHIPPER33 2020-05-14 16:34:02
Geometric Probability!!
Is Harvey here for geometric probability?
Equinox8 2020-05-14 16:34:02
Geometric probability
DPatrick 2020-05-14 16:34:19
Harvey's on break with rrusczyk, but fortunately I think (I hope!) I can handle geometric probability.
DPatrick 2020-05-14 16:34:29
Why does "geometric probability" feel like it'll be the right thing to try here?
walrus987 2020-05-14 16:35:08
because it's geometric and it's probability?
eibc 2020-05-14 16:35:08
because there are infintely many possibilities
FALCONSRULE1 2020-05-14 16:35:08
it is geometry and probabality
missionsqhc 2020-05-14 16:35:08
We have infinitely many points.
cubi 2020-05-14 16:35:08
can't count all the possible positions
DPatrick 2020-05-14 16:35:26
Right. There are infinitely many places the dart could land: any point on the dartboard is fair game!
DPatrick 2020-05-14 16:35:33
So we can't possibly hope to count them.
rayfish 2020-05-14 16:35:57
$\text{probability}=\frac{\text{area of successful region}}{\text{area of triangle}}$
aaravdodhia 2020-05-14 16:36:10
Probability = $\frac{\text{Successful Area}}{\text{Possible Area}}$.
subr3587835 2020-05-14 16:36:21
Area of success/total area
DPatrick 2020-05-14 16:36:23
Right! To compute the probability, we'll want to use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$
DPatrick 2020-05-14 16:36:35
So we need to figure out what those regions are.
mathuser18 2020-05-14 16:36:47
Can we draw a diagram?
srim1027 2020-05-14 16:36:47
draw a diagram?
ZippityA 2020-05-14 16:36:47
We should draw a diagram first.
MathHayden 2020-05-14 16:36:47
first we draw the dartboard
Raiders26 2020-05-14 16:36:47
draw an area of solution on the dartboard
DPatrick 2020-05-14 16:37:05
Yep, we're definitely going to need to draw a picture!
Pokemon2 2020-05-14 16:37:13
could we coordinate bash?
CHIPPER33 2020-05-14 16:37:27
Create a coordinate system with the triangle, There are infinite points to calculate
Puffer13 2020-05-14 16:37:37
set the triangle on the coordinate plane
mjl09 2020-05-14 16:37:48
coordinates system
FALCONSRULE1 2020-05-14 16:37:48
coordinate plane
DPatrick 2020-05-14 16:37:58
Good idea. We're going to need to compute some distances, so setting this up on the coordinate plane may be helpful.
DPatrick 2020-05-14 16:38:21
Any suggestions how to set it up?
Imishkabob 2020-05-14 16:38:46
0,0 could be one point
fasterthanlight 2020-05-14 16:38:46
Set one corner to be (0,0)
Jomo 2020-05-14 16:38:46
set center of base at origin
Asterlan 2020-05-14 16:38:46
One corner 0,0
KRISHIV 2020-05-14 16:38:46
set one point on (0,0) and work from there
DPatrick 2020-05-14 16:38:56
The origin is a really convenient point of the coordinate plane!
DPatrick 2020-05-14 16:39:06
So let's put the base of the dartboard along the $x$-axis, so that one corner is at $(0,0)$ and one corner is at $(6,0)$:
DPatrick 2020-05-14 16:39:12
DPatrick 2020-05-14 16:39:16
What point is the top of the dartboard?
AlexWangMath 2020-05-14 16:39:38
3-4-5 triangles
Math5K 2020-05-14 16:39:38
triangle vertices: (0,0),(3,4),(6,0)
kdraganov 2020-05-14 16:39:38
(0,0) -- (3,4) -- (0,6)
Wannabe 3-4-5 triangle
Set the points at $(0,0)$, $(6,0)$, and $(3,4)$.
awesomeming327. 2020-05-14 16:39:38
3,4
Feifei01 2020-05-14 16:39:38
(3, 4)
kdraganov 2020-05-14 16:39:38
$(3,4)$
asbodke 2020-05-14 16:39:38
(3,4)
AMC_Kid 2020-05-14 16:39:38
(3,4)
Math5K 2020-05-14 16:39:38
(3,4)
cai40 2020-05-14 16:39:38
(3,4)
bobamilkcha 2020-05-14 16:39:38
(3,4)
yayatheduck 2020-05-14 16:39:38
3,4
punkinpiday 2020-05-14 16:39:38
(3,4)
sixoneeight 2020-05-14 16:39:38
3,4
GrizzyProblemSolver79c 2020-05-14 16:39:38
(3, 4)
DPatrick 2020-05-14 16:39:50
Yes, 3-4-5 triangles FTW! Drawing the altitude of the dartboard to the side of length $6$ divides the board into two $3$-$4$-$5$ right triangles, so the top point is the point $(3,4)$:
DPatrick 2020-05-14 16:40:01
DPatrick 2020-05-14 16:40:33
So remember, what we're trying to do is use $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}}.$$
DPatrick 2020-05-14 16:40:43
Once of these is pretty easy.
Mrs. H math teacher 2020-05-14 16:41:05
Area of the triangle is 12.
happyhari 2020-05-14 16:41:05
the denominator is 12
ssr_07 2020-05-14 16:41:05
So height is 4 and base is 6. The area is 12 units squared.
Chesssaga 2020-05-14 16:41:05
Area of possible region = 12
cat_maniac_8 2020-05-14 16:41:05
12
rneelam 2020-05-14 16:41:05
its 12
MathBluebird 2020-05-14 16:41:05
area of triangle = 12
Wontoflonto 2020-05-14 16:41:05
area of the possible region, 12
cj13609517288 2020-05-14 16:41:05
possible region is just 12
MathIsFun286 2020-05-14 16:41:05
area of possible region is $12$
song2sons 2020-05-14 16:41:05
area of possible region is 12
millburn2006 2020-05-14 16:41:05
area of possible region = 12
DPatrick 2020-05-14 16:41:29
The "possible region" is the entire dartboard, because that's where the dart could land.
DPatrick 2020-05-14 16:41:38
So the denominator of our probability is just the area of the triangle.
DPatrick 2020-05-14 16:41:46
Since the base has length $6$ and the height is $4,$ its area is $\frac12(6)(4) = 12.$
DPatrick 2020-05-14 16:41:56
Now for the harder part. How do we determine the area of the "successful" region?
winterrain01 2020-05-14 16:42:30
Use the distance formula?
manokid 2020-05-14 16:42:30
Distance formula
ant08 2020-05-14 16:42:30
distance formula
vsamc 2020-05-14 16:42:30
Pick an arbitrary point $(a,b)$ inside the triangle
CHIPPER33 2020-05-14 16:42:30
Call the random point (x,y) and add up the distance squared to vertexes.
dudavid 2020-05-14 16:42:30
Distance formula
Meyzeek_Saveer 2020-05-14 16:42:30
distance formula
DPatrick 2020-05-14 16:42:48
Right! Suppose the dart lands at some point $(x,y).$
DPatrick 2020-05-14 16:42:59
We can write an expression for the sum of the squares of the distances from the dart to the corners of the board, using the distance formula.
DPatrick 2020-05-14 16:43:04
What's the square of the distance from $(x,y)$ to $(0,0)?$
Maths4J 2020-05-14 16:43:24
x^2+y^2
Imayormaynotknowcalculus 2020-05-14 16:43:24
$x^2+y^2$
TheEpicCarrot7 2020-05-14 16:43:24
x^2+y^2
Alculator11 2020-05-14 16:43:24
$x^2+y^2$
Speedstorm 2020-05-14 16:43:24
$x^2+y^2$
HappySpring 2020-05-14 16:43:24
sqrt (x^2 + y^2)
Jerry_Guo 2020-05-14 16:43:24
x^2+y^2
scoutskylar 2020-05-14 16:43:24
$x^2+y^2$
yukrant1 2020-05-14 16:43:24
$x^2+y^2$
DPatrick 2020-05-14 16:43:31
It's $x^2 + y^2.$
DPatrick 2020-05-14 16:43:34
What's the square of the distance from $(x,y)$ to $(6,0)?$
doglover07 2020-05-14 16:43:54
(x-6)^2+y^2
Leonard_my_dude 2020-05-14 16:43:54
(x-6)^2 + y^2
melonlord 2020-05-14 16:43:54
(6-x)^2 +y^2
Gumball 2020-05-14 16:43:54
(x-6)^2+y^2
nathanqiu 2020-05-14 16:43:54
(x-6)^2+y^2
mathcounts_genius 2020-05-14 16:43:54
(x-6)^2+y^2
Shmileyface 2020-05-14 16:43:54
(x-6)^2 + y^2
SharonW 2020-05-14 16:43:54
(x-6)^2+y^2
Puffer13 2020-05-14 16:43:54
(x-6)^2+y^2
DPatrick 2020-05-14 16:44:00
It's $(x-6)^2 + y^2.$
DPatrick 2020-05-14 16:44:04
What's the square of the distance from $(x,y)$ to $(3,4)?$
huela 2020-05-14 16:44:24
(x-3)^2+(y-4)^2
lrjr24 2020-05-14 16:44:24
$(x-3)^2+(y-4)^2$
menlo 2020-05-14 16:44:24
(x-3)^2 + (y-4)^2
Kruxe 2020-05-14 16:44:24
(x - 3)^2 + (y - 4)^2
C0atimundi 2020-05-14 16:44:24
(x-3)^2 * (y-4)^2
walrus987 2020-05-14 16:44:24
(x-3)^2 + (y-4)^2
ellnoo 2020-05-14 16:44:24
(x-3)^2+(y-4)^2
srim1027 2020-05-14 16:44:24
(x-3)^2 + (y-4)^2
NerdyDude 2020-05-14 16:44:24
(x-3)^2+(y-4)^2
jai123 2020-05-14 16:44:24
(x-3)^2+(y-4)^2
DPatrick 2020-05-14 16:44:30
It's $(x-3)^2 + (y-4)^2.$
DPatrick 2020-05-14 16:44:37
And for the dart to be successful, we need the sum of these to be less than $30$.
DPatrick 2020-05-14 16:44:41
That is, we need $$\left(x^2+y^2\right) + \left((x-6)^2+y^2\right) + \left((x-3)^2+(y-4)^2\right) < 30.$$
DPatrick 2020-05-14 16:44:50
Oof, what a mess. What should we do now?
Ninja-Girl 2020-05-14 16:45:15
It looks like a circle!
aidni47 2020-05-14 16:45:15
that's in the form of the equation for a circle!
That’s a circle
DPatrick 2020-05-14 16:45:30
It does look vaguely like the equation for a circle.
DPatrick 2020-05-14 16:45:55
And we shouldn't be too surprised if the question expects us to have $\pi$ in our answer.
DPatrick 2020-05-14 16:46:13
So we want to make this inequality look more like the equation for a circle.
EricShi1685 2020-05-14 16:46:26
simplify
nascar48 2020-05-14 16:46:26
expand
Shmileyface 2020-05-14 16:46:26
expand
menlo 2020-05-14 16:46:26
combine like terms
universeking 2020-05-14 16:46:26
expand!
DonutHole 2020-05-14 16:46:26
expand
fasterthanlight 2020-05-14 16:46:26
expand
Asterlan 2020-05-14 16:46:26
Expand
Kruxe 2020-05-14 16:46:26
combine like terms?
bedwinprusik578 2020-05-14 16:46:26
expand
sixoneeight 2020-05-14 16:46:26
Expand?
look4pilot 2020-05-14 16:46:26
expand?
DPatrick 2020-05-14 16:46:48
Sure, let's clean it up by expanding all the squares and combining like terms.
DPatrick 2020-05-14 16:46:57
I'll save us a little time: expanding and simplifying the left-hand side gives $3x^2 -18x +3y^2 -8y + 61 < 30.$
DPatrick 2020-05-14 16:47:19
Hmmm...still doesn't look quite like a circle to me.
DPatrick 2020-05-14 16:47:32
A circle's equation looks like $$(x-h)^2 + (y-k)^2 = r^2.$$
tumbleweed 2020-05-14 16:47:47
complete the square
Puffer13 2020-05-14 16:47:47
complete the square
EricShi1685 2020-05-14 16:47:47
complete the square
Alculator11 2020-05-14 16:47:47
Now, let's complete the square
Math5K 2020-05-14 16:47:47
complete the square
menlo 2020-05-14 16:47:47
complete the square!!!!
MathNerd555 2020-05-14 16:47:47
Completing the square?
Leonard_my_dude 2020-05-14 16:47:47
complete the square
FearlessTaurus 2020-05-14 16:47:47
Complete squares
happyhari 2020-05-14 16:47:47
complete the square now
cat_maniac_8 2020-05-14 16:47:47
Complete the square?
bobamilkcha 2020-05-14 16:47:47
complete the square
DPatrick 2020-05-14 16:47:55
Aha, you said the magic words!
DPatrick 2020-05-14 16:47:59
We need to complete the square in both $x$ and $y.$
ant08 2020-05-14 16:48:06
divdide by 3
DarthMaul 2020-05-14 16:48:06
subtract 61
edjar 2020-05-14 16:48:06
subtract the 61
Ilikeminecraft 2020-05-14 16:48:06
divide by 3
Lux1 2020-05-14 16:48:06
now we can subtract 61 from both sides
Ninja11 2020-05-14 16:48:09
divide by 3
rjiangbz 2020-05-14 16:48:09
divide by 3
jef23 2020-05-14 16:48:09
Subtract 61 from both sides
DPatrick 2020-05-14 16:48:21
Good idea: let's clean it up a bit more first by subtracting $61$ from both sides and dividing both sides by $3,$ to get $x^2 -6x + y^2 - \frac83y < -\frac{31}{3}.$
DPatrick 2020-05-14 16:48:37
Now what do we need to do to complete the square for $x$?
sosiaops 2020-05-14 16:48:56
+9
evanshawn316 2020-05-14 16:48:56
agentmath 2020-05-14 16:48:56
PureSwag 2020-05-14 16:48:56
add $9$
KevinW12345 2020-05-14 16:48:56
xMidnightFirex 2020-05-14 16:48:56
asbodke 2020-05-14 16:48:56
MathIsFun286 2020-05-14 16:48:56
add $9$
pikachu5829 2020-05-14 16:48:56
+9
Zhaom 2020-05-14 16:48:56
AMC_Kid 2020-05-14 16:48:56
DPatrick 2020-05-14 16:49:00
We need to add $9$ to complete the square for $x$, since $x^2 - 6x + 9 = (x-3)^2.$
DPatrick 2020-05-14 16:49:07
And what about for $y$?
Chesssaga 2020-05-14 16:49:22
Heavytoothpaste 2020-05-14 16:49:22
Jomo 2020-05-14 16:49:22
kdraganov 2020-05-14 16:49:22
add $\dfrac{16}{9}$
MR_67 2020-05-14 16:49:22
+16/9
RedFireTruck 2020-05-14 16:49:22
AmpharosX 2020-05-14 16:49:22
+16/9
CHIPPER33 2020-05-14 16:49:22
Dalar25 2020-05-14 16:49:22
+16/9
genius_007 2020-05-14 16:49:22
DPatrick 2020-05-14 16:49:27
We need to add $\dfrac{16}{9}$ to complete the square for $y,$ since $y^2 - \dfrac83y + \dfrac{16}{9} = \left(y - \dfrac43\right)^2.$
DPatrick 2020-05-14 16:49:35
And of course, if we add these to the left side of the inequality, we have to add them to the right side too!
DPatrick 2020-05-14 16:49:43
So the right side becomes $-\dfrac{31}{3} + 9 + \dfrac{16}{9} = \dfrac{-93+81+16}{9} = \dfrac49.$
DPatrick 2020-05-14 16:49:57
Therefore, completing the square in $x$ and $y$ gives$(x-3)^2 + \left(y-\frac43\right)^2 < \frac49.$
DPatrick 2020-05-14 16:50:09
Great! And what is this region?
IAmTheHazard 2020-05-14 16:50:36
It's a circle with radius $\frac{2}{3}$!
winterrain01 2020-05-14 16:50:36
A circle centered at (3, 4/3) with radius 2/3
KevinW12345 2020-05-14 16:50:36
a circle with center (3, 4/3)
jacobwu 2020-05-14 16:50:36
yukrant1 2020-05-14 16:50:36
A circle with radius 2/3 and center (3,4/3)
Puffer13 2020-05-14 16:50:36
a circle centered at (3,4/3) with radius 2/3
ryanfu2008 2020-05-14 16:50:36
menlo 2020-05-14 16:50:36
center at (3,4/3)
phoenixtan 2020-05-14 16:50:36
center is 3,4/3, radius is 2/3
Pokemon2 2020-05-14 16:50:36
that's a circle with center (3,4/3) and radius 2/3
DPatrick 2020-05-14 16:50:43
If this were an equation instead of an inequality, it would describe a circle with center $\left(3,\dfrac43\right)$ and radius $\sqrt{\dfrac49} = \dfrac23.$
Eth007 2020-05-14 16:50:54
the inside of a circle
Facejo 2020-05-14 16:51:00
Inside a circle with radius $\frac{2}{3}$
DPatrick 2020-05-14 16:51:03
So the inequality describes the interior of that circle, which is entirely inside the triangular dartboard.
DPatrick 2020-05-14 16:51:06
EulerRocks2.718 2020-05-14 16:51:32
Area of Success
dineshs 2020-05-14 16:51:32
disc with area 4pi/9
mathfun42 2020-05-14 16:51:32
a circle with area 4pi/9
casp 2020-05-14 16:51:32
4pi/9
NonOxyCrustacean 2020-05-14 16:51:32
4/9pi
cai40 2020-05-14 16:51:32
so the successful area is 4/9pi
Mandakini 2020-05-14 16:51:32
now just pi r^2
vsurya 2020-05-14 16:51:32
4/9 pi is area
DPatrick 2020-05-14 16:51:37
It's area is $\pi r^2 = \pi\left(\dfrac23\right)^2 = \dfrac49\pi.$
DPatrick 2020-05-14 16:51:43
That's the area of our "successful region".
DPatrick 2020-05-14 16:51:49
And to finish, what is the probability?
mmjguitar 2020-05-14 16:52:11
pi/27
ellnoo 2020-05-14 16:52:11
pi/27
donguri 2020-05-14 16:52:11
So the answer is $\frac{\pi }{27}$
slodha 2020-05-14 16:52:11
4/9 pi / 12
CT17 2020-05-14 16:52:11
$\frac{\frac{4}{9}\pi}{12}=\frac{\pi}{27}$
yizhou76 2020-05-14 16:52:11
millburn2006 2020-05-14 16:52:11
pi/27
Eng123 2020-05-14 16:52:11
$\pi/27$
onedance 2020-05-14 16:52:11
so it is pi/27
andrewhaitian 2020-05-14 16:52:11
pi/27
slodha 2020-05-14 16:52:11
4 / 9 pi / 12 is the probability
ZOP123 2020-05-14 16:52:11
pi/27
DPatrick 2020-05-14 16:52:14
It is $$P(\text{success}) = \dfrac{\text{Area of successful region}}{\text{Area of possible region}} =\dfrac{\frac49\pi}{12}.$$
DPatrick 2020-05-14 16:52:36
And this simplifies to our final answer of $\boxed{\frac{\pi}{27}}.$
DPatrick 2020-05-14 16:52:57
OK, back to rrusczyk! I'll be back later.
Meyzeek_Saveer 2020-05-14 16:53:22
RICHARD MY MANNNNN
Puffer13 2020-05-14 16:53:22
HI RICHARD RUSCZYK
Speedstorm 2020-05-14 16:53:22
Hi Mr. Rusczyk
KRISHIV 2020-05-14 16:53:22
hi richard!!
pi_is_3.14 2020-05-14 16:53:22
hi rruscyzk
rrusczyk 2020-05-14 16:53:24
Hi!
tacos4life123 2020-05-14 16:53:28
wait why are we changing back and fourth
rrusczyk 2020-05-14 16:53:44
Because Dave and I are old, and get tired trying to keep up with all of you.
DPatrick 2020-05-14 16:53:55
Speak for yourself.
rponda 2020-05-14 16:53:59
You're not that old!
Pythongoras3 2020-05-14 16:54:22
how 'bout a break for us
rrusczyk 2020-05-14 16:54:31
Y'all don't look like you need a break to me!
rrusczyk 2020-05-14 16:54:43
You might after the next problem...
rrusczyk 2020-05-14 16:54:45
Sprint #30: Hank builds an increasing sequence of positive integers as follows: The first term is $1$ and the second term is $2.$ Each subsequent term is the smallest positive integer that does NOT form a three-term arithmetic sequence with any previous terms of the sequence. The first five terms of Hank's sequence are $1, 2, 4, 5, 10.$ How many of the first $729$ positive integers are terms in Hank's sequence?
sri1priya 2020-05-14 16:55:03
i loved this problem!!
rrusczyk 2020-05-14 16:55:06
Me, too!
rrusczyk 2020-05-14 16:55:07
Before we get too deep: is there anything suspicious in the problem statement?
edjar 2020-05-14 16:55:54
729
tenebrine 2020-05-14 16:55:54
729 = 3^6
walrus987 2020-05-14 16:55:54
729?
AGENT99 2020-05-14 16:55:54
729
mathlete5451006 2020-05-14 16:55:54
729
raghavendrapv 2020-05-14 16:55:54
Why 729?
$729=3^6$
Flameslinger001 2020-05-14 16:55:54
why $729$
Himank_Chhaya_3 2020-05-14 16:55:54
The 729
missionsqhc 2020-05-14 16:55:54
729 = 3 ^6
Noam2007 2020-05-14 16:55:54
Yes! $729=9^3=3^6.$
rrusczyk 2020-05-14 16:55:58
The number $729$ is a little suspicious to me.
rrusczyk 2020-05-14 16:55:59
After all, $729 = 3^6$ is a nice perfect power.
rrusczyk 2020-05-14 16:56:21
So, we might have our eyes peeled for powers of 3.
rrusczyk 2020-05-14 16:56:42
Where should we start?
evanshawn316 2020-05-14 16:56:54
patterns?
Bluejay24 2020-05-14 16:56:54
the sequence?????
MTHJJS 2020-05-14 16:56:54
lets find a pattern!
Wontoflonto 2020-05-14 16:56:54
find a pattern?
rrusczyk 2020-05-14 16:57:00
Let's look for some patterns.
rrusczyk 2020-05-14 16:57:37
I see some of you asking what the question means. Let's figure that out by trying to build Hank's sequence. Maybe that will help us figure out what's going on this problem.
rrusczyk 2020-05-14 16:57:54
This "messing around" is a very powerful problem-solving strategy.
rrusczyk 2020-05-14 16:58:18
All right, let's start from the beginning:
rrusczyk 2020-05-14 16:58:23
rrusczyk 2020-05-14 16:58:32
We include $1$.
MLiang2018 2020-05-14 16:58:45
2
MLiang2018 2020-05-14 16:58:45
2
gs_2006 2020-05-14 16:58:45
next is 2
MathNerd555 2020-05-14 16:58:45
and 2
xMidnightFirex 2020-05-14 16:58:45
and 2
akpi2 2020-05-14 16:58:52
second 2
Jing.Han 2020-05-14 16:58:52
2
cai40 2020-05-14 16:58:52
the second term is 2
RedFireTruck 2020-05-14 16:58:52
2
rrusczyk 2020-05-14 16:59:00
Next is 2:
rrusczyk 2020-05-14 16:59:04
rrusczyk 2020-05-14 16:59:11
Can we include the 3?
qianqian07 2020-05-14 16:59:31
no
EricShi1685 2020-05-14 16:59:31
no
moony_eyed 2020-05-14 16:59:31
no
doudou_boston 2020-05-14 16:59:31
no
yesufsa 2020-05-14 16:59:31
No
mathissocool 2020-05-14 16:59:31
no
FALCONSRULE1 2020-05-14 16:59:31
no
vtlev 2020-05-14 16:59:31
no
FearlessTaurus 2020-05-14 16:59:31
no
Math_Genius_beast.com 2020-05-14 16:59:31
no
yayatheduck 2020-05-14 16:59:31
no
Ilikeminecraft 2020-05-14 16:59:31
no
Eng123 2020-05-14 16:59:31
NOOOOO.
EulerRocks2.718 2020-05-14 16:59:31
no
rneelam 2020-05-14 16:59:31
no
BurgerKingFootMan 2020-05-14 16:59:31
no
rrusczyk 2020-05-14 16:59:42
There were about 300 more no's in response...
rrusczyk 2020-05-14 16:59:48
Why can't we include the 3?
jacobwu 2020-05-14 17:00:00
NO that would make a sequence
CHIPPER33 2020-05-14 17:00:00
NO! 1,2,3 is an arithmetic sequence
christopherp 2020-05-14 17:00:00
no, 1-2-3 is an arithmetic sequence
ant08 2020-05-14 17:00:00
no, 1+2=3
Wontoflonto 2020-05-14 17:00:00
nah, 1-2-3 is a sequence
No since 1, 2, 3 is an arithmetic sequence.
rrusczyk 2020-05-14 17:00:41
Right -- 1,2,3 makes an arithmetic sequence: we add the same thing to get from the first to the second term as we do to get from the second to the third.
rrusczyk 2020-05-14 17:00:43
So, no 3:
rrusczyk 2020-05-14 17:00:46
MLiang2018 2020-05-14 17:01:03
so 4
Somersett 2020-05-14 17:01:03
Next is 4
Ninja11 2020-05-14 17:01:03
we can include 4
Alculator11 2020-05-14 17:01:03
But we can have 4
ElNoraa 2020-05-14 17:01:03
4
akpi 2020-05-14 17:01:03
so 4
Mandakini 2020-05-14 17:01:03
then 4
Meyzeek_Saveer 2020-05-14 17:01:03
yes 4
romani 2020-05-14 17:01:03
then 4
cheerupaops 2020-05-14 17:01:03
4
Moonshine-Dragonwing 2020-05-14 17:01:03
next term is 4
Poki 2020-05-14 17:01:03
4 would work
Lux1 2020-05-14 17:01:03
then we have 4
dolphin7 2020-05-14 17:01:03
we can do 4
rrusczyk 2020-05-14 17:01:08
$4$ is a keeper:
rrusczyk 2020-05-14 17:01:12
Supernova283 2020-05-14 17:01:25
Put in 5
eagle702 2020-05-14 17:01:25
Then 5
s0arbeacon 2020-05-14 17:01:25
than 5
superagh 2020-05-14 17:01:25
5
karthic7073 2020-05-14 17:01:25
And then 5
Kruxe 2020-05-14 17:01:25
then 5 is next
rrusczyk 2020-05-14 17:01:45
Before we go circling $5,$ are there any numbers we know we can cross out now that we have included $4?$
Flameslinger001 2020-05-14 17:02:25
7
sarahAops2020 2020-05-14 17:02:25
7
fasterthanlight 2020-05-14 17:02:25
6
donguri 2020-05-14 17:02:25
6 and 7
DZL1 2020-05-14 17:02:25
6 and 7
punkinpiday 2020-05-14 17:02:25
6 and 7
rjiangbz 2020-05-14 17:02:25
6,7
pi_is_3.14 2020-05-14 17:02:25
6,7
enkitty 2020-05-14 17:02:25
We can't have 6 because of 4-5-6 and 2-4-6
ASMSRJ 2020-05-14 17:02:25
6,7
blueocean2019 2020-05-14 17:02:25
we can cross out 6
srim1027 2020-05-14 17:02:25
6, 7
mathcounts_genius 2020-05-14 17:02:25
7
Flameslinger001 2020-05-14 17:02:25
6,7
Math_Penguin123 2020-05-14 17:02:25
7 and 6
Thegreatboy90 2020-05-14 17:02:25
6 and 7
AwesomeLife_Math 2020-05-14 17:02:25
7 and 6.
rrusczyk 2020-05-14 17:02:30
We have the sequences $1$-$4$-$7$ and $2$-$4$-$6$.
rrusczyk 2020-05-14 17:02:34
So $6$ and $7$ get crossed out:
rrusczyk 2020-05-14 17:02:37
rrusczyk 2020-05-14 17:03:21
This is pretty sweet. Once we find a new number, we can immediately cross others out.
tumbleweed 2020-05-14 17:03:32
we can still have 5
Doudou_Chen 2020-05-14 17:03:32
and 5 gets circled?
Unicornzrock 2020-05-14 17:03:32
5 can be circleds
ZippityA 2020-05-14 17:03:32
5 is okay to circle.
Gumball 2020-05-14 17:03:35
we should circle 5
superagh 2020-05-14 17:03:39
circle 5
rrusczyk 2020-05-14 17:04:12
Now, we know that $5$ is a keeper because we have already eliminated all the ones that must be eliminated when we keep $1,2,$ and $4.$
rrusczyk 2020-05-14 17:04:21
$5$ is the smallest unused number, and since we haven't crossed it out, it's not part of an arithmetic sequence. So we circle it.
rrusczyk 2020-05-14 17:04:23
menlo 2020-05-14 17:04:58
8 & 9 cross them out
Math4Life7 2020-05-14 17:04:58
now we can also cross out 8 and 9
kdraganov 2020-05-14 17:04:58
cross out 8 and 9 then
Juneybug 2020-05-14 17:04:58
so that eliminates 8 and 9 for us
Knightofaops 2020-05-14 17:04:58
cross out 8,9
JQWERTY6 2020-05-14 17:04:58
not nine.... 1-5-9 is a sequence
sixoneeight 2020-05-14 17:04:58
2-5-8, so 8 is crossed out, and 1-5-9, so 9 is crossed out.
pandax2007 2020-05-14 17:04:58
8 doesnt work because 2-5-8
Windigo 2020-05-14 17:04:58
no 8 or 9 because 1-5-9 and 2-5-8
MuffledPie 2020-05-14 17:04:58
then no 8 or 9
mathfun42 2020-05-14 17:04:58
8 and 9 can be eliminated
MathIsFun286 2020-05-14 17:04:58
cross out 8
evanshawn316 2020-05-14 17:04:58
we can eliminate 8 now that we have 5
slodha 2020-05-14 17:04:58
remove 9 and 8
Quaoar 2020-05-14 17:04:58
That eliminates 8 and 9
btc433 2020-05-14 17:04:58
8 cannot work because 2,5,8 is an arithmetic sequence now that 5 is circled.
KingRavi 2020-05-14 17:04:58
cross out 8 and 9
mfro24 2020-05-14 17:04:58
now cross out 8 and 9
jai123 2020-05-14 17:04:58
then take out 8 and 9 1-5-9 and 2-5-8
Makemakika1 2020-05-14 17:04:58
No 8 (2, 5, 8), no 9 (1, 5, 9)
agentmath 2020-05-14 17:04:58
8 doesn't work because 2, 5, 8, 9 doesn't work because of 1, 5, 9
LovingPilot 2020-05-14 17:04:58
not 8 and 9, 2-5-8, 1-5,9
rrusczyk 2020-05-14 17:05:20
The sequences with 5 in the middle are:

1-5-9

2-5-8

4-5-6
rrusczyk 2020-05-14 17:05:23
So we have to cross out $8$ and $9$. ($6$ is already crossed-out.)
rrusczyk 2020-05-14 17:05:24
rrusczyk 2020-05-14 17:06:04
This is fun -- it's a little like the Sieve of Eratosthenes for finding primes!
doglover07 2020-05-14 17:06:26
circle 10!
HarleyMathCounts 2020-05-14 17:06:26
circle 10
kkomma10 2020-05-14 17:06:26
you can do 10
Heavytoothpaste 2020-05-14 17:06:26
10 works
mathking999 2020-05-14 17:06:26
We can circle 10.
smartatmath 2020-05-14 17:06:26
10 circle
palindrome868 2020-05-14 17:06:26
Chesssaga 2020-05-14 17:06:26
10 is a keeper
Anish.A 2020-05-14 17:06:26
10 does work
artemispi 2020-05-14 17:06:26
circle 10
ttmmrryy 2020-05-14 17:06:26
next is 10!!
gs_2006 2020-05-14 17:06:26
then circle the 10
anishcool11 2020-05-14 17:06:26
circle 10
rrusczyk 2020-05-14 17:06:31
We circle $10$, the smallest unmarked numbered:
rrusczyk 2020-05-14 17:06:32
Lux1 2020-05-14 17:07:16
and cross out 19, 18, 16, 15
zhuqingzhang 2020-05-14 17:07:16
once we put in 10, we can eliminate 15,16,18,and 19
nduracell 2020-05-14 17:07:16
Once we circle 10, we can cross off 15, 16, 18 and 19
jbear911 2020-05-14 17:07:16
Cross out 15 because 5-10-15
mathfun42 2020-05-14 17:07:16
You can cross out 15, 16, 18, and 19
edjar 2020-05-14 17:07:16
Cross out 15,16,18,19
bluemathcounts 2020-05-14 17:07:16
cross out 15,16,18, and 19
flissyquokka17 2020-05-14 17:07:16
then eliminate 19, 18, 16, 15: (1, 10, 19), (2, 10, 18), (4, 10, 16), (5, 10, 15)
bluemathcounts 2020-05-14 17:07:16
cross out 15,16,18, and 19.
cj13609517288 2020-05-14 17:07:16
15,16,18,19 are out.
melonlord 2020-05-14 17:07:16
cross out 15,16,18, and 19
coldcrazylogic 2020-05-14 17:07:16
15, 16, 18, and 19 are gone because of this
nduracell 2020-05-14 17:07:16
We can cross off 15, 16 18 and 19
rrusczyk 2020-05-14 17:07:40
All the numbers smaller than $10$ can form a sequence with $10$ like so:

1-10-19

2-10-18

4-10-16

5-10-15
rrusczyk 2020-05-14 17:07:42
So $15$, $16$, $18$, and $19$ all get crossed out.
rrusczyk 2020-05-14 17:07:44
nascar48 2020-05-14 17:08:01
11 works
broham 2020-05-14 17:08:01
11 works
sarahAops2020 2020-05-14 17:08:01
11
truffle 2020-05-14 17:08:01
11 works too
LlamaWarrior 2020-05-14 17:08:01
Circle 11 as well
Unicorn78 2020-05-14 17:08:01
I think 11 works too
halphaboys 2020-05-14 17:08:01
circle 11
Poki 2020-05-14 17:08:05
then 11 would work also
rrusczyk 2020-05-14 17:08:21
We circle 11, and we cross out:

1-11-21

2-11-20

4-11-18

5-11-17

10-11-12
rrusczyk 2020-05-14 17:08:23
Facejo 2020-05-14 17:08:49
circle $13$
kkomma10 2020-05-14 17:08:49
you can do 13
Then 13
donguri 2020-05-14 17:08:49
circle 13
MTHJJS 2020-05-14 17:08:49
and 13 works
sdattilo2002 2020-05-14 17:08:49
13 works
Thegreatboy90 2020-05-14 17:08:49
13 is next
MathNerd555 2020-05-14 17:08:49
13 works
MathMaster029 2020-05-14 17:08:49
13 works, circle it
rrusczyk 2020-05-14 17:08:53
We can continue like this.
rrusczyk 2020-05-14 17:08:54
$13$ gets circled, and $25$, $24$, $22$, $21$ will get crossed out.
rrusczyk 2020-05-14 17:08:55
Layla2018 2020-05-14 17:09:18
circle 14
kittysnowball43 2020-05-14 17:09:18
14 works too
melonlord 2020-05-14 17:09:18
crcle 14
ElNoraa 2020-05-14 17:09:18
14
donguri 2020-05-14 17:09:18
circle 14
bot101 2020-05-14 17:09:18
14
yesufsa 2020-05-14 17:09:18
14 can be circled
walrus987 2020-05-14 17:09:18
14 can stay
winterrain01 2020-05-14 17:09:18
Circle 14
happyhari 2020-05-14 17:09:18
then 14
danprathab 2020-05-14 17:09:18
Circle 14
The_math_kid 2020-05-14 17:09:18
14 works
rrusczyk 2020-05-14 17:09:20
$14$ gets circled, and $27$, $26$, and $23$ get crossed out.
rrusczyk 2020-05-14 17:09:21
rrusczyk 2020-05-14 17:09:35
So the next number in the sequence is $28$.
jef23 2020-05-14 17:09:59
Except we can’t just go on like this till 729...
doglover07 2020-05-14 17:09:59
this is gonna take forever
rrusczyk 2020-05-14 17:10:11
Yeah, this is gonna take a while if we keep going...
rrusczyk 2020-05-14 17:10:30
Now, back at the beginning, what did we decide we were going to look out for?
FearlessTaurus 2020-05-14 17:10:47
powers of 3
aidni47 2020-05-14 17:10:47
powers of 3
Green4Applez 2020-05-14 17:10:47
powers of 3
BurgerKingFootMan 2020-05-14 17:10:47
powers of 3!
menlo 2020-05-14 17:10:47
multiles/powers of 3
scoutskylar 2020-05-14 17:10:47
powers of 3
onedance 2020-05-14 17:10:50
powers of 3
harmonyguan 2020-05-14 17:10:50
powers of 3
rrusczyk 2020-05-14 17:10:57
All right, let's look at our grid.
rrusczyk 2020-05-14 17:11:13
AGENT99 2020-05-14 17:11:25
All the numbers after powers of 3 work
vtlev 2020-05-14 17:11:29
10=3^2+1 and 28=3^3+1
cj13609517288 2020-05-14 17:11:32
10=3^2+1 and 28=3^3+1
rrusczyk 2020-05-14 17:11:40
Oh, that's interesting!
punkinpiday 2020-05-14 17:12:02
2 of the first 3, 4 of the first 9, 8 of the first 27.. 64 of the first 729!
john0512 2020-05-14 17:12:02
2 circled to 3, 4 circled to 9, 8 circled to 27
scoutskylar 2020-05-14 17:12:14
Up to 3^1, there are 2^1 numbers in the sequence. Up to 3^2, there are 2^2. Up to 3^3, there are 2^3. This pattern continues: up to 3^n, there are 2^n numbers in the sequence.
rrusczyk 2020-05-14 17:12:17
Ah, that's interesting, too!
samsonrao 2020-05-14 17:12:54
2 numbers before 3, 4 numbers before 9, 8 numbers before 27. Maybe 2^n numbers before 3^n
rrusczyk 2020-05-14 17:13:08
So, we see something interesting in our powers of $3.$ And some of us are seeing powers of $2,$ too. Let's stick with the powers of $3$ first.
rrusczyk 2020-05-14 17:13:34
How might we look at each integer if we are thinking about powers of $3?$
put them in base 3
Speedstorm 2020-05-14 17:13:55
Base 3!
CHIPPER33 2020-05-14 17:13:55
base 3
pi_is_3.14 2020-05-14 17:13:55
base 3
dolphin7 2020-05-14 17:13:55
base 3
amuthupss 2020-05-14 17:13:55
in base 3?
Speedstorm 2020-05-14 17:13:55
base 3
rrusczyk 2020-05-14 17:14:10
Lot's of stuff about powers of $3,$ let's think in base $3.$
rrusczyk 2020-05-14 17:14:20
Hank's sequence so far then would be:
rrusczyk 2020-05-14 17:14:28
$$1_3, 2_3, 11_3, 12_3, 101_3, 102_3, 111_3, 112_3, 1001_3.$$
mathking999 2020-05-14 17:14:47
All the ending units digits are one or two
Robot7620_2 2020-05-14 17:15:07
everything ends in 1 or 2
donguri 2020-05-14 17:15:07
Units digit are either 1 or 2
Jomo 2020-05-14 17:15:07
All the ones that have first digits of 1s and last digits of 1 or 2
rrusczyk 2020-05-14 17:15:11
Indeed, the final digit is always $1$ or $2.$
rrusczyk 2020-05-14 17:15:15
And the other digits?
edjar 2020-05-14 17:15:37
and all the rest are 0 or 1
mfro24 2020-05-14 17:15:37
First two digits are all 0 or 1
mfro24 2020-05-14 17:15:37
always 1 or 0
Superior 2020-05-14 17:15:37
1 and 0s
Gentoo 2020-05-14 17:15:37
1 or 0
edjar 2020-05-14 17:15:37
middles are 0 or 1
Alculator11 2020-05-14 17:15:39
The other digits are 1 or 0
ant08 2020-05-14 17:15:42
1 or 0
egerns 2020-05-14 17:15:42
0 or 1
olive0827 2020-05-14 17:15:42
just 1 or 0
nkt2002 2020-05-14 17:15:42
1 or 0
rrusczyk 2020-05-14 17:15:56
Just $0$ and $1$ for the other digits...
scoutskylar 2020-05-14 17:16:23
There aren't any digit 2s if you subtract 1!
rrusczyk 2020-05-14 17:16:45
Oh, very interesting! Let's subtract $1$ from every number. Then we get:
rrusczyk 2020-05-14 17:16:50
$$0_3, 1_3, 10_3, 11_3, 100_3, 101_3, 110_3, 111_3, 1000_3.$$
scinderella220 2020-05-14 17:17:00
Wow
Quaoar 2020-05-14 17:17:00
Woah.
danprathab 2020-05-14 17:17:27
WOW
agentmath 2020-05-14 17:17:27
wow
winterrain01 2020-05-14 17:17:27
Those look like binary
LlamaWarrior 2020-05-14 17:17:27
Amazing
Imayormaynotknowcalculus 2020-05-14 17:17:27
Pretty cool
zhuqingzhang 2020-05-14 17:17:27
Wow
PShucks 2020-05-14 17:17:27
WOah1
RedFireTruck 2020-05-14 17:17:27
thats binary
r0518 2020-05-14 17:17:27
now we only have 0 and 1
sri1priya 2020-05-14 17:17:27
it becomes binary!
EulerRocks2.718 2020-05-14 17:17:27
Binary
Ajtyb12dfj 2020-05-14 17:17:27
wow!
MathBluebird 2020-05-14 17:17:27
That's so cool!
PShucks 2020-05-14 17:17:27
WOOAH!
rrusczyk 2020-05-14 17:17:56
Yeah, that's just plain pretty. Let's just look at it for a little while.
scoutskylar 2020-05-14 17:18:05
That isn't binary.
rrusczyk 2020-05-14 17:18:22
Correct -- we aren't looking at this in base 2 right now.
aidni47 2020-05-14 17:18:27
its in base 3
KRISHIV 2020-05-14 17:18:33
yea its in base 3 still
jef23 2020-05-14 17:18:33
Yeah bc it’s in base 3
BurgerKingFootMan 2020-05-14 17:18:33
that's base 3
rrusczyk 2020-05-14 17:18:36
It is in base 3.
rrusczyk 2020-05-14 17:18:39
But...
dineshs 2020-05-14 17:18:41
It looks like binary
rrusczyk 2020-05-14 17:18:53
It sure looks like binary.
fasterthanlight 2020-05-14 17:19:17
think of it as binary
NHatAOPS 2020-05-14 17:19:17
if it was binary, it would be 1 2 3 4 5...
floatmeeting 2020-05-14 17:19:17
It's every number in binary too
rrusczyk 2020-05-14 17:20:06
If we just think of these as being binary numbers.... WE HAVE ALL THE POSITIVE INTEGERS. Which makes counting the number of terms in Hank's sequence up through a given number really easy.
nathanqiu 2020-05-14 17:20:13
AND THAT EXPLAINS THE 2^N! BECAUSE THERE ARE 2^N NUMBERS OF N DIGITS IN BINARY!!!
Neberg 2020-05-14 17:20:21
There are two options for each digit, 0 or 1, so there are $2^n$ numbers less than $3^n$ (because there are 2 options for each of n digits).
rrusczyk 2020-05-14 17:20:22
Exactly.
rrusczyk 2020-05-14 17:20:45
So, we have a guess now for our answer.
MuffledPie 2020-05-14 17:21:09
729=3^6,so 2^6=64
Ninja11 2020-05-14 17:21:09
2^6 is 64
Chesssaga 2020-05-14 17:21:09
DottedCaculator 2020-05-14 17:21:09
64
cooljoseph 2020-05-14 17:21:09
64
try11out 2020-05-14 17:21:09
64
Windigo 2020-05-14 17:21:09
64
NHatAOPS 2020-05-14 17:21:09
64
MathWizard09 2020-05-14 17:21:09
64
manokid 2020-05-14 17:21:09
64
MathJams 2020-05-14 17:21:09
64?
rrusczyk 2020-05-14 17:21:26
We think that there are $64$ terms in Hank's sequence.
rrusczyk 2020-05-14 17:21:33
Let's recap why we believe this.
rrusczyk 2020-05-14 17:21:38
Since $729 = 1000000_3,$ the numbers from $0$ to $728$ are all the numbers we can write in base $3$ with $6$ or fewer digits.
rrusczyk 2020-05-14 17:22:19
But in our list, we want only the numbers of $0$ or $1$ as digits. (Including our leading $0$s).
rrusczyk 2020-05-14 17:23:00
So, we have $6$ slots to fill (again, including possibly with those leading $0$s), and each slot has $2$ choices: $0$ or $1.$
skyleristhecoolest 2020-05-14 17:23:05
so 2^6
AbhiWwis 2020-05-14 17:23:05
2^6
rrusczyk 2020-05-14 17:23:31
So, there are $2^6$ of these base-$3$ numbers less than $729$ with only $0$ and $1$ as digits.
rrusczyk 2020-05-14 17:23:47
(That's another way to look at the observation we made by reading everything as binary.)
sri1priya 2020-05-14 17:23:59
that makes this problem the coolest!
rrusczyk 2020-05-14 17:24:02
Indeed.
rrusczyk 2020-05-14 17:24:04
But.
Scipow 2020-05-14 17:24:18
but why does this happen?
cj13609517288 2020-05-14 17:24:18
Unfortunately, we need to prove this conjecture.
Chesssaga 2020-05-14 17:24:27
we still need to prove it rigorously
amuthupss 2020-05-14 17:24:29
we need to prove it works
MR_67 2020-05-14 17:24:42
how do we prove this
RedFireTruck 2020-05-14 17:24:42
we need to use a proof
rrusczyk 2020-05-14 17:24:46
Yeah, I'm not satisfied either.
rrusczyk 2020-05-14 17:24:55
I think we have the answer.
rrusczyk 2020-05-14 17:25:03
But as a mathematician, I want to try to prove that the conjecture is true.
rrusczyk 2020-05-14 17:25:23
(I might not do this during the test. But afterwards, wouldn't be able to sleep until I figured it out.)
Zoobat 2020-05-14 17:25:30
How do you do this in less than 2 minutes?
rrusczyk 2020-05-14 17:25:45
See the pattern, hope it continues, write down $64$, prove it later.
rneelam 2020-05-14 17:25:53
are we gonna prove it now
rrusczyk 2020-05-14 17:25:57
We're gonna try.
rrusczyk 2020-05-14 17:26:01
There are two steps: we have to prove that the sequence we have satisfies the "no three terms form an arithmetic sequence" property, and we have to prove that we can't add any new terms to the sequence to make it bigger.
rrusczyk 2020-05-14 17:26:15
Let's do these one at a time.
rrusczyk 2020-05-14 17:26:30
First, we show that if we only have numbers with $0$s and $1$s as base-$3$ digits, then no three of them can form an arithmetic sequence.
rrusczyk 2020-05-14 17:26:37
Let's suppose we had three terms $x < y < z$. What equation would have to be true for this to be an arithmetic sequence?
Puffer13 2020-05-14 17:27:08
2y=x+z
rjiangbz 2020-05-14 17:27:08
x+z=2y
superagh 2020-05-14 17:27:08
2y=x+z
doglover07 2020-05-14 17:27:08
y-x=z-y
XTJin 2020-05-14 17:27:08
x+z=2y
Leonard_my_dude 2020-05-14 17:27:08
x+z=2y
BakedPotato66 2020-05-14 17:27:08
x+z=2y
cindyzou 2020-05-14 17:27:08
y-x = z-y
MY-2 2020-05-14 17:27:08
y-x = z-y
amuthupss 2020-05-14 17:27:08
y-x = z-y
edjar 2020-05-14 17:27:08
y-x=z-y
ant08 2020-05-14 17:27:08
x+z=2y
MathNerd555 2020-05-14 17:27:08
y-x=z-y
rrusczyk 2020-05-14 17:27:31
We would need $y$ to be the average of $x$ and $z$. (Or we can start from $y-x = z-y,$ as many of you nicely did.)
rrusczyk 2020-05-14 17:27:36
The simplest way to write this is $2y = x+z$.
rrusczyk 2020-05-14 17:28:56
Now, if $x, y,$ and $z$ are in our sequence written in base $3$ (after we subtracted $1$), then what do we know about that right-hand side $x+z?$
donguri 2020-05-14 17:29:08
There are no carries?
edjar 2020-05-14 17:29:08
No carrying
acornfirst 2020-05-14 17:29:20
it would have no carries
rrusczyk 2020-05-14 17:29:51
For each digit in the sum, we add a $0$ or $1$ to another $0$ or $1.$
awesomebooks 2020-05-14 17:29:57
no carries
CHIPPER33 2020-05-14 17:29:57
no carries
john0512 2020-05-14 17:29:57
No carrying
happyhari 2020-05-14 17:29:57
no carries
mohanty 2020-05-14 17:29:57
no carry over
NO CARRIES
rrusczyk 2020-05-14 17:30:01
So, no carries.
bishope6 2020-05-14 17:30:09
It could end in a 0, 1, or 2.
Green4Applez 2020-05-14 17:30:17
it will have digits of 0, 1, and 2
Maths4J 2020-05-14 17:30:17
It has 0s, 1s, and 2s in it
rrusczyk 2020-05-14 17:30:30
Each digit will be $0,1,$ or $2.$ Hmmm....
rrusczyk 2020-05-14 17:30:38
Maybe that's not so interesting. Let's look at the other side.
rrusczyk 2020-05-14 17:30:44
What will the left side look like?
Leonard_my_dude 2020-05-14 17:31:11
also no carries
Chesssaga 2020-05-14 17:31:11
0 or 2
mfro24 2020-05-14 17:31:11
all 0s or 2s
eez 2020-05-14 17:31:11
a series of 0s and 2s
CHIPPER33 2020-05-14 17:31:11
end in 0 or 2
AGENT99 2020-05-14 17:31:11
0 or 2
scoutskylar 2020-05-14 17:31:11
there can't be any 1s.
cj13609517288 2020-05-14 17:31:11
Each digit can only be 0 or 2.
Eng123 2020-05-14 17:31:11
0 or 2
MathIsFun286 2020-05-14 17:31:11
no carries
FireDragon1719 2020-05-14 17:31:13
it will be either a $0,2$
rrusczyk 2020-05-14 17:31:18
Also no carries.
rrusczyk 2020-05-14 17:31:32
$y$ only has $0$ and $1$ as digits, so we double $y$ by doubling each digit.
rrusczyk 2020-05-14 17:31:47
But there are no $1$s in $2y.$
rrusczyk 2020-05-14 17:32:02
Looking back over at the right hand side....
rrusczyk 2020-05-14 17:32:29
Is it possible that the right-hand side will not have any $1$s?
edjar 2020-05-14 17:32:59
And in the original there has to be at least 1 one
astumbur 2020-05-14 17:32:59
no
PShucks 2020-05-14 17:32:59
no
maxben 2020-05-14 17:32:59
no
no
PShucks 2020-05-14 17:32:59
no!
cat_maniac_8 2020-05-14 17:32:59
No
m_goli 2020-05-14 17:32:59
no
DottedCaculator 2020-05-14 17:32:59
no
PShucks 2020-05-14 17:32:59
no!
MathWizard09 2020-05-14 17:32:59
no
rrusczyk 2020-05-14 17:33:06
Why not?
asbodke 2020-05-14 17:33:16
Only if they are the same number
nathanqiu 2020-05-14 17:33:16
only if there is no 1+0 or 0+1
Gentoo 2020-05-14 17:33:16
Each 1 will have to match up with another 1
mohanty 2020-05-14 17:33:33
there has to be at least one 1
there has to be at least one 1
rrusczyk 2020-05-14 17:33:42
Ah! $x$ and $z$ are different!
rrusczyk 2020-05-14 17:33:54
Some digit of our sum $x+z$ will come from adding a $0$ from $x$ and a $1$ for $z,$ or the reverse.
There has to be at least 1
rrusczyk 2020-05-14 17:34:09
So, we have to get a $1$ somewhere!
yukrant1 2020-05-14 17:34:31
then 2y can never equal x+z
rrusczyk 2020-05-14 17:34:37
So, $2y$ has no $1$s. But $x+z$ has at least one $1$.
Puffer13 2020-05-14 17:34:41
so they cannot be equal
rrusczyk 2020-05-14 17:35:03
So we cannot have $2y = x + z$. And thus there's no arithmetic sequence in our sequence that has only those numbers with $0$s and $1$s.
cj13609517288 2020-05-14 17:35:09
And we are done!!

Oh wait, that's only half of the proof.
rrusczyk 2020-05-14 17:35:45
Fortunately, our work so far is most of the proof of the second part of the conjecture: why can't we include any numbers with a $2$ as a digit in the sequence?
rrusczyk 2020-05-14 17:35:48
Let's try an example. Suppose we tried to insert $z = 120221_3$ into our sequence as we're building it. Remember, we've already included all the smaller numbers in the sequence that consist only of $0$s and $1$s in base $3.$
rrusczyk 2020-05-14 17:36:06
Can you find an $x$ and a $y$ already in the sequence -- so that $x$ and $y$ only has $0$s and $1$s -- so that $x,y,z$ are an arithmetic sequence?
EFrame2 2020-05-14 17:37:34
100001 110111
edjar 2020-05-14 17:37:41
x=100001 y= 110111
huela 2020-05-14 17:37:42
100001, 110111, 120221
rrusczyk 2020-05-14 17:37:52
Just take $x$ to be the same number as $z$, but with the $2$s replaced by $0$s.
rrusczyk 2020-05-14 17:37:55
So $x = 100001_3.$
rrusczyk 2020-05-14 17:37:59
And what happens when we add $x+z?$
sixoneeight 2020-05-14 17:38:25
you get only 2s
MathBluebird 2020-05-14 17:38:25
we get 2y
sixoneeight 2020-05-14 17:38:25
2s and 0s only.
Puffer13 2020-05-14 17:38:25
220222
huela 2020-05-14 17:38:25
220222
AGENT99 2020-05-14 17:38:25
220222
kdraganov 2020-05-14 17:38:25
$220222_3$
fasterthanlight 2020-05-14 17:38:27
220222
Lux1 2020-05-14 17:38:28
the y has only 2s and 0s
rrusczyk 2020-05-14 17:38:33
Again, there's no carrying! If $z$ has a $0$ or $1,$ then $x$ does too, but if $z$ has a $2,$ then $x$ has a $0.$
rrusczyk 2020-05-14 17:38:34
And furthermore, all the digit sums are $0$s or $2$s! In our example $x+z = 220222_3.$
rrusczyk 2020-05-14 17:38:35
That means we can divide by $2!$
rrusczyk 2020-05-14 17:38:37
All the $2$s become $1$s, and we have $y = 110111_3$ with all $0$s and $1$s.
rrusczyk 2020-05-14 17:39:25
This process generalizes! Given any $z$, take $x$ to be the same as $z$ but with its $2$s replaced by $0$s. Then $x+z$ will have only $0$s and $2$s, so $y = \frac12(x+z)$ will have only $0$s and $1$s, and $x,y,z$ is an arithmetic sequence. (Notice that $y$ is what you get when you replace the $2$s in $z$s with $1$s.)
Puffer13 2020-05-14 17:39:36
yaaah
RedPenMan 2020-05-14 17:39:36
woah
cj13609517288 2020-05-14 17:39:36
Wait, but that's it!
sdattilo2002 2020-05-14 17:39:36
cool!
edjar 2020-05-14 17:39:40
We can do this with any number so our proof is complete and the answer to the problem is 64
EulerRocks2.718 2020-05-14 17:39:47
Cool!
MathIsFun286 2020-05-14 17:39:47
wow
Math5K 2020-05-14 17:39:47
wow!
danprathab 2020-05-14 17:39:49
AND WE ARE DONE WITH OUR PROOF ! ! !
which proves that 64 or 2^6 is the answer!
Lux1 2020-05-14 17:39:55
wow. this problem is neat!
CHIPPER33 2020-05-14 17:39:55
That's amazing!
PShucks 2020-05-14 17:39:55
yay
missionsqhc 2020-05-14 17:39:55
nice
PShucks 2020-05-14 17:39:55
wow
yay
NinjaMango 2020-05-14 17:40:01
Wow!
jupiter314 2020-05-14 17:40:01
great!
aidni47 2020-05-14 17:40:01
wow!
mfro24 2020-05-14 17:40:01
That was awesome!
JBB 2020-05-14 17:40:01
wow
donguri 2020-05-14 17:40:01
YAY!
akpi 2020-05-14 17:40:01
yay
Tiger2010 2020-05-14 17:40:01
brilliant
nathanqiu 2020-05-14 17:40:04
AND WE'RE DONE!!!
doglover07 2020-05-14 17:40:04
yay
Chesssaga 2020-05-14 17:40:04
YAYAYAYAYAYAY
rrusczyk 2020-05-14 17:40:08
So we've proved the conjecture, and now proved that the answer is indeed $\boxed{64}$.
rrusczyk 2020-05-14 17:40:17
Yeah, that proof was pretty tough!
rrusczyk 2020-05-14 17:40:33
Totally cool if you didn't follow it -- we went through it a lot faster than it took me to find it
Jomo 2020-05-14 17:40:43
Are we going to do the target problems?
donguri 2020-05-14 17:40:43
Are we doing any target problems?
sosiaops 2020-05-14 17:40:46
are we doing target round?
rrusczyk 2020-05-14 17:40:56
If by "we" you mean "DPatrick", why, yes!
DPatrick 2020-05-14 17:41:10
OK, now you can get your calculator out. We're switching to Target Round problems!
onedance 2020-05-14 17:41:29
dun dun dun!
Kush0924 2020-05-14 17:41:29
Target is the best
evanshawn316 2020-05-14 17:41:29
AWESOME!
DPatrick 2020-05-14 17:41:37
Target #6: Iris is playing a game that has a $5 \times 5$ gameboard like the one shown. The goal is to get her game piece from the square labeled $\star$ to the square labeled $\circ$ using a series of moves any positive integer number of squares up or any positive integer number of squares to the right. Note that moving two squares up in a single move is different than moving two squares up in two moves. How many unique sequences of moves can Iris make to get her game piece from $\star$ to $\circ$?
DPatrick 2020-05-14 17:41:43
DPatrick 2020-05-14 17:42:09
Any suggestions?
Leonard_my_dude 2020-05-14 17:42:50
count the number of ways to each square
yayy 2020-05-14 17:42:50
count the ways to get to every square
Asterlan 2020-05-14 17:42:50
write the number of ways to get to each square in them
Makorn 2020-05-14 17:43:45
start filling squares in with how many ways u can get from the starred square to that square; u can calculate this by summing the ways from all possible previous squares
austinchen2005 2020-05-14 17:43:45
you can fill in squares with number of ways, and you figure out how many number of ways on a square from the sum of the squares to the left and below it
bishope6 2020-05-14 17:43:59
Start by figuring out how many sequences of moves there are to get to the star from the squares nearest to the star, and work backwards.
DPatrick 2020-05-14 17:44:26
I think this is a good general strategy: we'll try to count the number of paths to or from each square.
DPatrick 2020-05-14 17:44:47
It doesn't really matter too much whether we start at the star and count forward, or start at the circle and count backwards.
DPatrick 2020-05-14 17:45:25
But because I wrote my diagrams for the second way, let's work backwards. (It's pretty much the same count.)
DPatrick 2020-05-14 17:45:38
That is, we start at the $\circ$ and count backwards how many ways to get to $\circ$ for each space.
MathWizard38025 2020-05-14 17:45:50
but wont that take a really long time
DPatrick 2020-05-14 17:46:00
I hope not! The grid isn't too big. Let's see what happens.
DPatrick 2020-05-14 17:46:11
We start at $\circ$. There's only $1$ way to get to $\circ$ from $\circ$--you're already there!
DPatrick 2020-05-14 17:46:15
What about the two spaces immediately adjacent to $\circ$?
rjiangbz 2020-05-14 17:46:31
1
Math5K 2020-05-14 17:46:31
1
aidni47 2020-05-14 17:46:31
1
Superior 2020-05-14 17:46:31
1 way
vsurya 2020-05-14 17:46:31
1
SkywalkerAUV 2020-05-14 17:46:31
each 1
jbear911 2020-05-14 17:46:31
1
egerns 2020-05-14 17:46:31
1
EulerRocks2.718 2020-05-14 17:46:31
1 way
green_alligator 2020-05-14 17:46:31
1 way
bobthegod78 2020-05-14 17:46:31
1 way
MathIsFun286 2020-05-14 17:46:31
1 and 1
Maths4J 2020-05-14 17:46:31
both 1
DPatrick 2020-05-14 17:46:35
There's also only $1$ way to get to $\circ$ from each of the spaces adjacent to $\circ$.
DPatrick 2020-05-14 17:46:49
Let's keep track with a picture.
DPatrick 2020-05-14 17:46:55
DPatrick 2020-05-14 17:47:04
Next, let's consider the square with the red $X.$ How many ways to get to $\circ$ from $X?$
Ninja11 2020-05-14 17:47:22
x=2
nathanqiu 2020-05-14 17:47:22
X=2
winterrain01 2020-05-14 17:47:22
2
MathAMC8 2020-05-14 17:47:22
2 way
Eng123 2020-05-14 17:47:22
2
scoutskylar 2020-05-14 17:47:22
2
JBB 2020-05-14 17:47:22
2
eez 2020-05-14 17:47:22
2
Ninja11 2020-05-14 17:47:22
2
Mathemats 2020-05-14 17:47:22
2
SharonW 2020-05-14 17:47:22
2
yukrant1 2020-05-14 17:47:22
2
huela 2020-05-14 17:47:22
2
yayatheduck 2020-05-14 17:47:22
2
Sri_Math 2020-05-14 17:47:22
x=2
superagh 2020-05-14 17:47:22
2
DPatrick 2020-05-14 17:47:32
From there, we can take one step right (and then there is $1$ way to finish).

Or we can take one step up (and then there is $1$ way to finish).
DPatrick 2020-05-14 17:47:37
Combining these, we have $1+1=2$ ways to finish from $X.$
DPatrick 2020-05-14 17:47:44
DPatrick 2020-05-14 17:47:54
What about $Y$? How many ways to $\circ$ from $Y?$
ChrisalonaLiverspur 2020-05-14 17:48:08
$2$ ways.
Floating-Clouds 2020-05-14 17:48:08
2 ways.
akpi2 2020-05-14 17:48:08
2
PShucks 2020-05-14 17:48:08
2
mathgenius237 2020-05-14 17:48:08
2
MathWiz20 2020-05-14 17:48:08
2
PShucks 2020-05-14 17:48:08
2!
Poki 2020-05-14 17:48:08
2
PShucks 2020-05-14 17:48:08
2
m_goli 2020-05-14 17:48:08
2
albertedwin 2020-05-14 17:48:08
2
DPatrick 2020-05-14 17:48:15
We can take one step right, with $1$ way to finish from there.

Or, we can take two steps right, with $1$ way to finish from there--already done!
DPatrick 2020-05-14 17:48:21
Combining these, we have $1+1=2$ ways to finish from $Y.$
bobthegod78 2020-05-14 17:48:38
same for y and z
jupiter314 2020-05-14 17:48:38
y and z are also equal to 2
nathanqiu 2020-05-14 17:48:38
$Y,Z=2$
mfro24 2020-05-14 17:48:38
$2$ ways to $Z$ as well!
Z = 2 as well
Kruxe 2020-05-14 17:48:38
so there's 2 ways from Z as well
2
tumbleweed 2020-05-14 17:48:38
2
ARSM2019 2020-05-14 17:48:43
same thing for z right
DPatrick 2020-05-14 17:48:48
The situation at $Z$ is identical to that at $Y$, except that we are going up instead of to the right. We'll keep exploiting this symmetry throughout this solution!
DPatrick 2020-05-14 17:48:52
Let's update the diagram and label the next group of squares:
DPatrick 2020-05-14 17:48:57
DPatrick 2020-05-14 17:49:11
How many ways to finish from $A?$
Leonard_my_dude 2020-05-14 17:49:38
4
CrazyVideoGamez 2020-05-14 17:49:38
4
gs_2006 2020-05-14 17:49:38
4
ellnoo 2020-05-14 17:49:38
4
walrus987 2020-05-14 17:49:38
4
Jalenluorion 2020-05-14 17:49:38
4
nikenissan 2020-05-14 17:49:38
4 ways
JQWERTY6 2020-05-14 17:49:38
4
lihao_david 2020-05-14 17:49:38
4
RedFireTruck 2020-05-14 17:49:38
4
mfro24 2020-05-14 17:49:38
$A=4$
sdattilo2002 2020-05-14 17:49:38
4
sigma_notation 2020-05-14 17:49:38
4
aops_band 2020-05-14 17:49:38
4
Pianodude 2020-05-14 17:49:38
4
DPatrick 2020-05-14 17:49:43
We can jump from $A$ to any of the three squares to the right of $A.$
DPatrick 2020-05-14 17:49:55
And from those squares, we have $2$, $1$, and $1$ way to finish.
DPatrick 2020-05-14 17:50:00
So we have $2+1+1 = 4$ ways to get from $A$ to $\circ$.
mfro24 2020-05-14 17:50:13
$D=4$
sixoneeight 2020-05-14 17:50:13
D is 4 too!
jbear911 2020-05-14 17:50:13
same with D
ALDW123 2020-05-14 17:50:13
d=4 as well
Chesssaga 2020-05-14 17:50:13
Same with D
MathIsFun286 2020-05-14 17:50:13
$d=4$
winterrain01 2020-05-14 17:50:13
By symmetry, D is 4 too, right?
DPatrick 2020-05-14 17:50:26
Yep: by the symmetry of the grid, we have $4$ ways from $D$ too.
DPatrick 2020-05-14 17:50:30
And what about $B$?
Awesome3.14 2020-05-14 17:51:06
B=5 ways
TheEpicCarrot7 2020-05-14 17:51:06
B=5
Streaks123 2020-05-14 17:51:06
5
XTJin 2020-05-14 17:51:06
5
Awesome3.14 2020-05-14 17:51:06
5
lrjr24 2020-05-14 17:51:06
5.
Math4Life2020 2020-05-14 17:51:06
B = 5
john0512 2020-05-14 17:51:06
5
CHIPPER33 2020-05-14 17:51:06
b=5
sarahAops2020 2020-05-14 17:51:06
5
mfro24 2020-05-14 17:51:06
$B=5$
MR_67 2020-05-14 17:51:06
B and C = 5
CT17 2020-05-14 17:51:06
5
MrEgggga 2020-05-14 17:51:06
$B=5$
awesomeming327. 2020-05-14 17:51:06
5.
MR_67 2020-05-14 17:51:06
B=5
yayy 2020-05-14 17:51:06
5
Kason 2020-05-14 17:51:06
B is 5, 3+2, and C is by symmetry, 5,too
DPatrick 2020-05-14 17:51:10
From $B,$ we can go one step up ($2$ ways to finish), one step right ($2$ ways to finish), or two steps right ($1$ way to finish).
DPatrick 2020-05-14 17:51:18
So there are $2+2+1=5$ ways to finish from $B.$
DPatrick 2020-05-14 17:51:26
And the same for $C$, by symmetry.
DPatrick 2020-05-14 17:51:31
DPatrick 2020-05-14 17:51:39
Now what?
Math5K 2020-05-14 17:52:00
continue
MathIsFun286 2020-05-14 17:52:00
next diagonal
sixoneeight 2020-05-14 17:52:00
we keep going
CrazyVideoGamez 2020-05-14 17:52:00
Keep going on to the next squares!
nathanqiu 2020-05-14 17:52:03
this remeinds me of pascal's triangle
Jalenluorion 2020-05-14 17:52:20
looks a little like pascals riangle
sdattilo2002 2020-05-14 17:52:20
is this pascal's triangle in a weird way
DPatrick 2020-05-14 17:52:28
Right...it's a bit like Pascal's Triangle.
DPatrick 2020-05-14 17:52:59
Except that, instead of adding just the two adjacent numbers, we're adding all the numbers in the squares directly to the right of that square and all the numbers directly above that square.
walrus987 2020-05-14 17:53:07
it's pascal's except each thing is the sum of everything to the right and above, rather than the immediate right and above
DPatrick 2020-05-14 17:53:21
So let's work on the next diagonal:
DPatrick 2020-05-14 17:53:25
DPatrick 2020-05-14 17:53:33
Note that at this point I've built the symmetry into my labels: the two $E$s will be the same, as will the two $F$s.
ARSM2019 2020-05-14 17:53:55
all the way on the left is 8 then because 4+2+1+1=8
Kruxe 2020-05-14 17:53:55
so that means the top-left corner is 8?
Pianodude 2020-05-14 17:53:55
E=8
PShucks 2020-05-14 17:53:55
e=8
ARSM2019 2020-05-14 17:53:55
E=8
MR_67 2020-05-14 17:53:55
E = 8
menlo 2020-05-14 17:53:55
E=8
sri1priya 2020-05-14 17:53:55
e = 8
anishcool11 2020-05-14 17:53:55
E is 8
DarthMaul 2020-05-14 17:53:55
E: 8
Doudou_Chen 2020-05-14 17:53:55
E has 8
DPatrick 2020-05-14 17:54:00
From the top $E$, we sum the numbers to the right of $E$ and get $4+2+1+1 = 8.$
TheMagician 2020-05-14 17:54:26
f = 12
ccu700037 2020-05-14 17:54:26
F=12
mathbw225 2020-05-14 17:54:26
F=12
bobjoebilly 2020-05-14 17:54:26
F = 4 + 5 + 2 + 1 = 12
Dalar25 2020-05-14 17:54:26
F=12
Juneybug 2020-05-14 17:54:26
f=12
bronzetruck2016 2020-05-14 17:54:26
f=12
akpi2 2020-05-14 17:54:26
f=12
wertiol123 2020-05-14 17:54:26
F=12
mfro24 2020-05-14 17:54:26
$F = [4]+[1+2+5] = 12$
DPatrick 2020-05-14 17:54:30
From the top $F$, we sum the numbers above and to the right of $F$ and get $4+5+2+1 = 12.$
Math5K 2020-05-14 17:54:51
G=14
Dalar25 2020-05-14 17:54:51
G=14
G=14
mohanty 2020-05-14 17:54:51
G is 14
jjw4106 2020-05-14 17:54:51
G=14
blue_arrow 2020-05-14 17:54:51
G=14
aidni47 2020-05-14 17:54:51
g=14
EpicGirl 2020-05-14 17:54:51
G=14
MathBluebird 2020-05-14 17:54:51
g = 14
Math5K 2020-05-14 17:54:51
G=14.
Flying_Bird 2020-05-14 17:54:51
$G=14$
DPatrick 2020-05-14 17:54:55
From $G$, we sum the numbers above and to the right of $G$ and get $2+5+5+2 = 14.$
DPatrick 2020-05-14 17:54:59
DPatrick 2020-05-14 17:55:04
We keep going like this!
DPatrick 2020-05-14 17:55:07
What numbers are on the next diagonal?
albertedwin 2020-05-14 17:55:42
28
evanshawn316 2020-05-14 17:55:42
28
vcric247 2020-05-14 17:55:42
28
bobthegod78 2020-05-14 17:55:42
28, 37
Ninja11 2020-05-14 17:55:42
37
Lux1 2020-05-14 17:55:42
first one is 28
asbodke 2020-05-14 17:55:42
28 and 37
Maths4J 2020-05-14 17:55:42
28 and 37
fasterthanlight 2020-05-14 17:55:42
28,37,37,28
bot101 2020-05-14 17:55:42
28 and 37
vsurya 2020-05-14 17:55:42
28
AmpharosX 2020-05-14 17:55:42
28,37,37,28
evanshawn316 2020-05-14 17:55:42
28 and 37
pvmathnerd 2020-05-14 17:55:42
28 and 37
DPatrick 2020-05-14 17:55:46
The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$
DPatrick 2020-05-14 17:55:52
DPatrick 2020-05-14 17:55:54
And next?
sigma_notation 2020-05-14 17:56:27
94,106,94
Ninja-Girl 2020-05-14 17:56:27
94 106 94
mohanty 2020-05-14 17:56:27
94 and 106
Awesome3.14 2020-05-14 17:56:27
94 and 106
lihao_david 2020-05-14 17:56:27
94 and 106
Ninja11 2020-05-14 17:56:27
94 and 106
jupiter314 2020-05-14 17:56:27
94 and 106
Mathematicalbrain 2020-05-14 17:56:27
94 and 106
DPatrick 2020-05-14 17:56:48
Oops, I got out of sync by one diagonal!
DPatrick 2020-05-14 17:56:57
The prior diagonal was $8+12+5+2+1 = 28$ and $4+12+15+5+2 = 37.$
DPatrick 2020-05-14 17:57:04
The next diagonal has $8+28+37+14+5+2 = 94$ and $4+12+37+37+12+4 = 106.$
DPatrick 2020-05-14 17:57:08
DPatrick 2020-05-14 17:57:20
Getting near the finish line!
FearlessTaurus 2020-05-14 17:57:38
289 is next
bobjoebilly 2020-05-14 17:57:38
then 289
evanshawn316 2020-05-14 17:57:38
289
MR_67 2020-05-14 17:57:38
Next is 289 and 289
cat_maniac_8 2020-05-14 17:57:38
289 is the last ones!!!
Ninja11 2020-05-14 17:57:38
289
gordonhero 2020-05-14 17:57:38
289
yukrant1 2020-05-14 17:57:38
289,289 is next diagonal
TheEpicCarrot7 2020-05-14 17:57:38
289
DPatrick 2020-05-14 17:57:42
The next diagonal has $8+28+94+106+37+12+4 = 289.$
DPatrick 2020-05-14 17:57:56
DPatrick 2020-05-14 17:58:08
And the lower-left square, which is the final answer?
donguri 2020-05-14 17:58:28
The answer is $\boxed{838}$
CHIPPER33 2020-05-14 17:58:28
838
yayatheduck 2020-05-14 17:58:28
and finally 838!
CHIPPER33 2020-05-14 17:58:28
838!!!
Liakas_42 2020-05-14 17:58:28
838
Quaoar 2020-05-14 17:58:28
838
winterrain01 2020-05-14 17:58:28
lucaszhou 2020-05-14 17:58:28
838
nathanqiu 2020-05-14 17:58:28
838!!!
Sedro 2020-05-14 17:58:28
838
scoutskylar 2020-05-14 17:58:28
$\bbox[5px, border: 2px solid black]{838}$
concierge 2020-05-14 17:58:28
838
Zhaom 2020-05-14 17:58:28
838
so finnally, it is 838
DPatrick 2020-05-14 17:58:33
The final answer is that there are $8+28+94+289+289+94+28+8 = \boxed{838}$ paths from $\star$ to $\circ$.
DPatrick 2020-05-14 17:58:38
DPatrick 2020-05-14 17:59:06
Sometimes just getting your handy dirty and computing is the simplest way to the answer.
green_alligator 2020-05-14 17:59:26
is there a faster way to do it?
gs_2006 2020-05-14 17:59:26
is there an easier way or is this the only one
pianoboy 2020-05-14 17:59:26
that was long. How could I do that in a shorter way
DPatrick 2020-05-14 17:59:51
Not that I'm readily aware of. I don't really recognize the numbers we're seeing, so there's no obvious pattern that jumps out at me.
DPatrick 2020-05-14 18:00:12
OK, back to rrusczyk for some watermelon.
rrusczyk 2020-05-14 18:00:19
Yum!
rrusczyk 2020-05-14 18:00:22
Target #7: The figure shows a perfectly round watermelon, represented by circle $O,$ secured on a flat table, represented by segment $AB,$ using a rope. One end of the rope is attached to the table at $A,$ then draped over the top of the watermelon, and the other end is attached at $B.$ The watermelon has diameter $20$ cm. The center of the watermelon is $20$ cm from point $A$ and $10\sqrt{2}$ cm from point $B.$ What is the minimum length of rope necessary to secure the watermelon to the table? Express your answer as a decimal to the nearest tenth.
rrusczyk 2020-05-14 18:00:33
rrusczyk 2020-05-14 18:00:36
Just so we're clear on the problem: what we're trying to do is to measure the blue length shown:
rrusczyk 2020-05-14 18:00:40
rrusczyk 2020-05-14 18:00:51
Where should we start?
Jalenluorion 2020-05-14 18:01:23
draw a radius from o to ab
MathIsFun286 2020-05-14 18:01:23
radii to the points of tangency
SuperSine 2020-05-14 18:01:23
draw lines to the points of tangency
yayy 2020-05-14 18:01:27
draw the radii perpendicular to tangents
memi 2020-05-14 18:01:31
drawing radii to points of tangency
rrusczyk 2020-05-14 18:01:37
It's usually help to draw in radii for a circle where we think it will help.
rrusczyk 2020-05-14 18:01:42
Here, there are three radii just begging to be drawn -- radii to points of tangency! That will give us right angles, and we like right angles.
rrusczyk 2020-05-14 18:01:45
Let's draw the radii to the two points where the rope starts and stops running along the melon (and we'll call those points $C$ and $D$).
rrusczyk 2020-05-14 18:01:46
Let's also draw the radius to where the melon touches the table (and we'll call that point $E$).
rrusczyk 2020-05-14 18:01:54
And let's add the lengths we know to the picture. All the radii are $10$, and we have the two given lengths $OA = 20$ and $OB = 10\sqrt2.$
rrusczyk 2020-05-14 18:01:57
rrusczyk 2020-05-14 18:02:09
So our goal is to compute the total length $AC + \widehat{CD} + DB$, where $\widehat{CD}$ means the arc length from $C$ to $D.$
rrusczyk 2020-05-14 18:02:14
What's $AC?$
KingRavi 2020-05-14 18:02:42
AOC is 30 60 90 triangle
KingRavi 2020-05-14 18:02:42
10 sqrt 3
Derpy123 2020-05-14 18:02:42
10sqrt3
xMidnightFirex 2020-05-14 18:02:42
10sqrt3
SkywalkerAUV 2020-05-14 18:02:42
10sqrt3
Jalenluorion 2020-05-14 18:02:42
special right triangles!
harmonyguan 2020-05-14 18:02:42
10sqrt3
$10sqrt{3}$
Dalar25 2020-05-14 18:02:42
10*sqrt(3)
ARSM2019 2020-05-14 18:02:42
ac=10 root 3
Leonard_my_dude 2020-05-14 18:02:42
10sqrt3
Speedstorm 2020-05-14 18:02:42
$10\sqrt{3}$
Maths4J 2020-05-14 18:02:42
AC=10sqrt(3)
winner320 2020-05-14 18:02:42
10 root 3
LightiningBlazer 2020-05-14 18:02:42
10sqrt(3)
Math5K 2020-05-14 18:02:42
10sqrt(3)
asbodke 2020-05-14 18:02:42
$10\sqrt{3}$
Pianodude 2020-05-14 18:02:42
AC = 10sqrt3
Sri_Math 2020-05-14 18:02:42
10√3
aidni47 2020-05-14 18:02:42
10 rt 3
$10\sqrt{3}$
tumbleweed 2020-05-14 18:02:42
10\sqrt3
rrusczyk 2020-05-14 18:02:54
Side $OC$ is half of side $OA.$
rrusczyk 2020-05-14 18:02:57
So, $AOC$ is a 30-60-90 triangle!
rrusczyk 2020-05-14 18:02:59
So $AC = 10\sqrt3$.
rrusczyk 2020-05-14 18:03:00
How about $DB$?
xMidnightFirex 2020-05-14 18:03:48
we have square ODBE
Math5K 2020-05-14 18:03:48
30-60-90 and 45-45-90 triangles!
scoutskylar 2020-05-14 18:03:48
$10$
krishgarg 2020-05-14 18:03:48
10
CHIPPER33 2020-05-14 18:03:48
10, 45-45-90
vrondoS 2020-05-14 18:03:48
10
Significant 2020-05-14 18:03:48
10
sosiaops 2020-05-14 18:03:48
$10$
tientien1 2020-05-14 18:03:48
10
Jerry_Guo 2020-05-14 18:03:48
10
JunoSuno 2020-05-14 18:03:48
10
millburn2006 2020-05-14 18:03:48
10
punkinpiday 2020-05-14 18:03:48
DB = 10
akpi 2020-05-14 18:03:48
10
scinderella220 2020-05-14 18:03:48
10
Scifan671 2020-05-14 18:03:48
10
akpi 2020-05-14 18:03:48
it is 10
sosiaops 2020-05-14 18:03:48
$10$ !
Doudou_Chen 2020-05-14 18:03:48
45-45-90 triangle, so DB = 10!
Lux1 2020-05-14 18:03:48
it has a length of 10
KSH31415 2020-05-14 18:03:48
10 because it is a square
rrusczyk 2020-05-14 18:03:51
$ODB$ is a 45-45-90 triangle! (Or you might have seen that $ODBE$ is a square.)
rrusczyk 2020-05-14 18:03:52
So $DB = 10$.
rrusczyk 2020-05-14 18:03:53
And what about the arc $\widehat{CD}$?
Ninja11 2020-05-14 18:04:16
150 degress
Pokemon2 2020-05-14 18:04:16
150 degree measure?
harmonyguan 2020-05-14 18:04:16
150 degrees
Liakas_42 2020-05-14 18:04:16
150
winner320 2020-05-14 18:04:19
150 degree
MathIsFun286 2020-05-14 18:04:19
150 degrees
Patriots_Mathgeeks 2020-05-14 18:04:20
150 degress
rrusczyk 2020-05-14 18:04:34
How do we see that $m\angle COD$ is $150$ degrees?
DerpyTaterTot 2020-05-14 18:05:14
its 360-45-45-60-60
yayatheduck 2020-05-14 18:05:14
360-60-60-90
MR_67 2020-05-14 18:05:14
360-60-60-45-45
Jalenluorion 2020-05-14 18:05:14
360-60-60-45-45
KingRavi 2020-05-14 18:05:14
360-60-60-45-45
nathanqiu 2020-05-14 18:05:14
360-(60+60+45+45)
MathIsFun286 2020-05-14 18:05:14
360 -60-60-45-45=150
darksol 2020-05-14 18:05:14
360 - (2*45) - (2*60) = 150
sosiaops 2020-05-14 18:05:14
360-(60+60+45++45)
Awesome3.14 2020-05-14 18:05:14
360-90-60-60=150
sarasota25 2020-05-14 18:05:14
360-45-45-60-60=150
360-90-60-60
btc433 2020-05-14 18:05:17
$360-60-60-90=150$
rrusczyk 2020-05-14 18:05:22
We know that $m\angle COA = 60^\circ,$ and similarly $m\angle AOE = 60^\circ.$
rrusczyk 2020-05-14 18:05:23
And we also know that $m\angle DOE = 90^\circ$.
rrusczyk 2020-05-14 18:05:24
So we have $m\angle COD = 360^\circ - 90^\circ -60^\circ - 60^\circ = 150^\circ.$
rrusczyk 2020-05-14 18:05:42
All right, what's the length of arc $\widehat{CD}$?
CHIPPER33 2020-05-14 18:06:23
5/12 * circumference
sixoneeight 2020-05-14 18:06:23
so the length is 150/360*20pi=25/3*pi
Paulfrank 2020-05-14 18:06:23
100pi/12
tumbleweed 2020-05-14 18:06:23
$\frac{25\pi}{3}$
xMidnightFirex 2020-05-14 18:06:23
100pi/12 = 25pi/3
Lux1 2020-05-14 18:06:23
so then the arc would measure 15/36 of the circumference
Awesome3.14 2020-05-14 18:06:23
150/360 * 20pi
MR_67 2020-05-14 18:06:23
20pi(150/360)
IAM_TSSXII 2020-05-14 18:06:23
25 pi over 3
Xcountry 2020-05-14 18:06:23
we need 5/12 of the circumference
menlo 2020-05-14 18:06:23
150(100pi)/360
and.peggy 2020-05-14 18:06:23
(150/360)*2*10*pi
jai123 2020-05-14 18:06:23
150/360 times 20 pi
winterrain01 2020-05-14 18:06:23
150/360 * 20pi
rrusczyk 2020-05-14 18:06:26
We see that the curved portion of the rope is $\dfrac{150^\circ}{360^\circ} = \dfrac{5}{12}$ of the whole circumference of the watermelon.
rrusczyk 2020-05-14 18:06:27
Therefore, the curved portion of the rope has length $\dfrac{5}{12}(20\pi)$ cm.
rrusczyk 2020-05-14 18:06:32
Putting all the pieces of the rope together, its total length is $10\sqrt{3}+ \frac{5}{12}(20\pi) + 10.$
rrusczyk 2020-05-14 18:06:49
Time to get out your calculator!
onedance 2020-05-14 18:06:53
53.5?
CHIPPER33 2020-05-14 18:06:53
53.5!!
nathanqiu 2020-05-14 18:06:53
53.5!
Awesome3.14 2020-05-14 18:07:04
The length of the rope is 53.5
yukrant1 2020-05-14 18:07:04
53.5!!!
evanshawn316 2020-05-14 18:07:04
53.5
bronzetruck2016 2020-05-14 18:07:04
53.5
53.5
Heidi0902 2020-05-14 18:07:04
53.5
mohanty 2020-05-14 18:07:04
53.5
arirah9 2020-05-14 18:07:07
53.5!!!!!
Nefarious 2020-05-14 18:07:07
53.5
rrusczyk 2020-05-14 18:07:10
Compute this, and you should find that the rope is approximately $\boxed{53.5}$ cm long.
palindrome868 2020-05-14 18:07:15
yep!!
sosiaops 2020-05-14 18:07:15
WE DID IT!!!
sosiaops 2020-05-14 18:07:24
YAY WE DID IT!
YAY!
rrusczyk 2020-05-14 18:07:39
All right, back to DPatrick!
DPatrick 2020-05-14 18:07:47
Hi again!
DPatrick 2020-05-14 18:07:51
Target #8: Tyrell makes a list of the left-most digit of the powers of $2,$ from $2^0$ to $2^{1000},$ inclusive. The first six numbers on Tyrell’s list, therefore, are $1, 2, 4, 8, 1, 3$ and the last six numbers are $3, 6, 1, 2, 5, 1.$ Given that $2^{1000}$ has $302$ digits, and that $8$ appears $52$ times on Tyrell’s list, how many times does $9$ appear in his list?
DPatrick 2020-05-14 18:08:11
Of all 46 problems on all three contests, I think this one was my favorite.
DPatrick 2020-05-14 18:08:33
We're given a lot of information, but it's all a little hard to use.
DPatrick 2020-05-14 18:08:44
Any ideas?
Maths4J 2020-05-14 18:09:12
46 is the max score but there are 48 problems, 10+8+30
aie8920 2020-05-14 18:09:12
Aren't there 48 problems
Quaoar 2020-05-14 18:09:17
Weren't there 48 problems?
DPatrick 2020-05-14 18:09:28
Oops, you're right! This one is still my favorite, though.
romani 2020-05-14 18:09:53
There is a pattern
Parentcoach_Maths 2020-05-14 18:09:53
Let’s look for a pattern
PShucks 2020-05-14 18:09:53
find a pattern!
LlamaWarrior 2020-05-14 18:09:53
Find a pattern!
PShucks 2020-05-14 18:09:53
find a pattern
Unicorn78 2020-05-14 18:09:53
Look for patterns?
Leonard_my_dude 2020-05-14 18:09:53
List some out and hope it repeats i guess
mathfun42 2020-05-14 18:09:53
it forms a pattern
arirah9 2020-05-14 18:09:53
look for a pattern
MathyAOP 2020-05-14 18:09:53
Look for a pattern
Liakas_42 2020-05-14 18:09:53
find a pattern
find a pattern
akpi2 2020-05-14 18:09:53
patterns
winterrain01 2020-05-14 18:09:53
Just get in and mess around?
mathbw225 2020-05-14 18:09:53
look for a pattern
DPatrick 2020-05-14 18:10:01
Yeah, let's look for patterns.
DPatrick 2020-05-14 18:10:29
Let's think about how the first digit can change. That is, let's think about what can follow each of the numbers in our list.
DPatrick 2020-05-14 18:10:33
For example, what can follow a $1$ in the list?
pianoboy 2020-05-14 18:10:56
2,3
Speedstorm 2020-05-14 18:10:56
2 or 3
mfro24 2020-05-14 18:10:56
2 or 3
awesomeming327. 2020-05-14 18:10:56
2,3
john0512 2020-05-14 18:10:56
2 or 3
Math5K 2020-05-14 18:10:56
2 or 3
AmpharosX 2020-05-14 18:10:56
2 and 3
geodash2 2020-05-14 18:10:56
2 or 3
2 or a 3
Leonard_my_dude 2020-05-14 18:10:56
2 or 3
DottedCaculator 2020-05-14 18:10:56
2 or 3
MTHJJS 2020-05-14 18:10:56
2 or 3
yukrant1 2020-05-14 18:10:56
a 2 or a 3?
qsheng 2020-05-14 18:10:56
2 or 3
I-_-I 2020-05-14 18:10:56
2 or 3
DPatrick 2020-05-14 18:11:03
Right. If we double a number that starts with $1$, then the resulting number must start with $2$ or $3.$
DPatrick 2020-05-14 18:11:13
You might find thinking about small-length numbers helpful. If we double a number between 100 and 199, the resulting number is between 200 and 398. If we double a number between 1000 and 1999, the resulting number is between 2000 and 3998. And so on, regardless of the number of digits.
DPatrick 2020-05-14 18:11:25
So $1$ must always be followed by $2$ or $3.$
DPatrick 2020-05-14 18:11:37
How about $2$? What number(s) can follow $2$ in our list?
millburn2006 2020-05-14 18:11:54
4,5
fasterthanlight 2020-05-14 18:11:54
4,5
Math4Life2020 2020-05-14 18:11:54
4 5
jupiter314 2020-05-14 18:11:54
4 or 5
jbear911 2020-05-14 18:11:54
4 or 5
rjtmagic 2020-05-14 18:11:54
4/5
punkinpiday 2020-05-14 18:11:54
4 or 5
winner320 2020-05-14 18:11:54
2 must be followed ny 4 or 5
Ninja11 2020-05-14 18:11:54
4 or 5
kdraganov 2020-05-14 18:11:54
4 or 5
superagh 2020-05-14 18:11:54
4, 5
mathisfun17 2020-05-14 18:11:54
4,5
Chesssaga 2020-05-14 18:11:54
4,5
KSH31415 2020-05-14 18:11:54
4 and 5
casp 2020-05-14 18:11:54
4 or 5
DPatrick 2020-05-14 18:11:58
A $2$ must be followed by a $4$ or a $5$, by the same logic.
DPatrick 2020-05-14 18:12:13
How about $3$? What can follow $3$?
MathCounts145 2020-05-14 18:12:28
6 or 7
Jerry_Guo 2020-05-14 18:12:28
6 or 7
green_alligator 2020-05-14 18:12:28
6, 7
ElNoraa 2020-05-14 18:12:28
6,7
vcric247 2020-05-14 18:12:28
6/7
Asterlan 2020-05-14 18:12:28
6 or 7
ohbaby 2020-05-14 18:12:28
6 or 7
TheEpicCarrot7 2020-05-14 18:12:28
6 and 7
felicitas 2020-05-14 18:12:28
6 or 7
ccu700037 2020-05-14 18:12:28
6 or 7
jellybeanchocolate 2020-05-14 18:12:28
6 or 7
acornfirst 2020-05-14 18:12:28
6 or 7
DPatrick 2020-05-14 18:12:33
A $3$ must be followed by a $6$ or a $7.$
DPatrick 2020-05-14 18:12:38
How about $4$?
menlo 2020-05-14 18:12:53
8,9
krishgarg 2020-05-14 18:12:53
8,9
MathIsFun286 2020-05-14 18:12:53
8,9
QQMath 2020-05-14 18:12:53
8,9
gs_2006 2020-05-14 18:12:53
4 is an 8 or a 9
lucaszhou 2020-05-14 18:12:53
8,9
Gentoo 2020-05-14 18:12:53
8 or 9
pi_is_3.14 2020-05-14 18:12:53
8, 9
tumbleweed 2020-05-14 18:12:53
8,9
SR1234 2020-05-14 18:12:53
8 or 9
vitaRaptor2000 2020-05-14 18:12:53
8 or 9
UnicornForev19 2020-05-14 18:12:53
8 or 9
Floating-Clouds 2020-05-14 18:12:53
8 or 9
DPatrick 2020-05-14 18:12:59
A $4$ must be followed by an $8$ or a $9.$
DPatrick 2020-05-14 18:13:10
How about $5?$
Awesome3.14 2020-05-14 18:13:30
only 1
MR_67 2020-05-14 18:13:30
1
cat_maniac_8 2020-05-14 18:13:30
1
NCEE 2020-05-14 18:13:30
1
TThB0501 2020-05-14 18:13:30
1
Zhaom 2020-05-14 18:13:30
1 and only 1
TMSmathcounts737 2020-05-14 18:13:30
and 5 must be followed with a 1
Fishareweird 2020-05-14 18:13:30
1
NinjaMango 2020-05-14 18:13:30
1
FireCurve24 2020-05-14 18:13:30
1
DPatrick 2020-05-14 18:13:33
A $5$ is always followed by a $1.$ Why?
acornfirst 2020-05-14 18:14:17
because it carries over
Anniboy 2020-05-14 18:14:17
it is the leftmost digit
Lux1 2020-05-14 18:14:17
because 5*2 is 10
AmpharosX 2020-05-14 18:14:17
Because new digit!
$5\cdot2=10$
harmonyguan 2020-05-14 18:14:17
cuz 5*2=10
mmjguitar 2020-05-14 18:14:17
becase it is a new place value
Math5K 2020-05-14 18:14:17
50*2=100 and 59*2=118
axzhang 2020-05-14 18:14:17
because it will create a new digit infront
Juneybug 2020-05-14 18:14:17
oh yeah, because it turns into either a 10 or 11, both of which start with 1
Dalar25 2020-05-14 18:14:17
because if you double the number, it will come out to 10 or 11, both of which start with 1
E1CAarush 2020-05-14 18:14:17
11 and 10
DPatrick 2020-05-14 18:14:26
Right: doubling a number with a $5$ will add a digit! For example, doubling a number between $500$ and $599$ gives us a number between $1000$ and $1198$.
DPatrick 2020-05-14 18:14:47
So we always go to a number that's got one more digit, and the new number starts with $1$.
DPatrick 2020-05-14 18:14:57
So a $5$ in our list is always followed by a $1$.
sixoneeight 2020-05-14 18:15:11
5,6,7,8,9 all give 1!
Potato2017 2020-05-14 18:15:11
6 is 1, 7 is 1, 8 is 1, 9 is 1.
Pokemon2 2020-05-14 18:15:11
5-9 always gives a 1
Chesssaga 2020-05-14 18:15:11
Same goes for 6-9
Noam2007 2020-05-14 18:15:11
Same with $6,7,8,$ and $9$ because they all add a new digit.
sixoneeight 2020-05-14 18:15:11
so are 6,7,8, and 9!
DPatrick 2020-05-14 18:15:32
Aha! It's the same with $6, 7, 8,$ and $9$: they always are followed by a $1.$
DPatrick 2020-05-14 18:15:47
So we keep cycling back to $1$, every time we add a digit to the underlying number (the power of $2$).
yukrant1 2020-05-14 18:15:54
and going back to 1 restarts the pattern
Noam2007 2020-05-14 18:16:03
That means that $1$'s are the most common.
DPatrick 2020-05-14 18:16:10
Aha...$1$'s are the key.
DPatrick 2020-05-14 18:16:31
We have a $1$ in our list, and we double for a few steps, and we eventually add a digit to our power of $2$ which gets us back to a new $1$ in our list.
E1CAarush 2020-05-14 18:16:44
We need to find a pattern with ones
DPatrick 2020-05-14 18:17:00
I agree. What are the different possible cycles from one $1$ to the next $1$?
qsheng 2020-05-14 18:17:27
1,2,4,8,1
MR_67 2020-05-14 18:17:27
1,2,4,8,1
Zhaom 2020-05-14 18:17:27
1-2-4-8-1
Jerry_Guo 2020-05-14 18:17:27
1, 2, 4, 8, 1
bobjoebilly 2020-05-14 18:17:27
1,2,4,8,1
casp 2020-05-14 18:17:27
1,2,4,8,1
mmjguitar 2020-05-14 18:17:27
1,2,4,8,1
bacond 2020-05-14 18:17:27
1,2,4,8,1
DPatrick 2020-05-14 18:17:31
We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$. (Indeed, this is how the list starts.)
DPatrick 2020-05-14 18:17:39
What else?
Doudou_Chen 2020-05-14 18:17:55
1-2-4-9-1-
Lcz 2020-05-14 18:18:03
1 2 4 9 1
mathfun42 2020-05-14 18:18:03
1, 2, 4, 9, 1
DPatrick 2020-05-14 18:18:06
We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.
maxben 2020-05-14 18:18:16
1 2 5 1
1-2-5-1
eibc 2020-05-14 18:18:16
1, 2, 5, 1
Eng123 2020-05-14 18:18:16
1, 2, 5, 1
Speedstorm 2020-05-14 18:18:16
1,2,5,1
DPatrick 2020-05-14 18:18:20
We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.
sarahAops2020 2020-05-14 18:18:29
1-3-6-1
Jalenluorion 2020-05-14 18:18:29
1361
aidni47 2020-05-14 18:18:29
1, 3, 6, 1
CHIPPER33 2020-05-14 18:18:29
1,3,6,1
albertzhuan 2020-05-14 18:18:29
1, 3, 6, 1
ApraTrip 2020-05-14 18:18:29
1,3,6,1
manokid 2020-05-14 18:18:29
1361
DPatrick 2020-05-14 18:18:33
We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.
superagh 2020-05-14 18:18:54
1 3 7 1
palindrome868 2020-05-14 18:18:54
1-3-7-1
nathanqiu 2020-05-14 18:18:54
1,3,7,1
onedance 2020-05-14 18:18:54
1,3,7,1
sdattilo2002 2020-05-14 18:18:54
1,3,7,1
DPatrick 2020-05-14 18:18:57
We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.
DPatrick 2020-05-14 18:19:21
And that's it: if you look at the rules we discovered for what numbers can follow which, there's are the only cycles from one $1$ to the next.
DPatrick 2020-05-14 18:19:29
Let me list them all out again:
DPatrick 2020-05-14 18:19:44
We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 8 \rightarrow 1$.
We can go $1 \rightarrow 2 \rightarrow 4 \rightarrow 9 \rightarrow 1$.
We can go $1 \rightarrow 2 \rightarrow 5 \rightarrow 1$.
We can go $1 \rightarrow 3 \rightarrow 6 \rightarrow 1$.
We can go $1 \rightarrow 3 \rightarrow 7 \rightarrow 1$.
XTJin 2020-05-14 18:20:04
8 only appears in one of the paths... 9 only appears in one of the paths as well...
asbodke 2020-05-14 18:20:12
1 8 and 1 9 in total
Maths4J 2020-05-14 18:20:12
So the sum of all the 8s and 9s equals the number of 4s.
bedwinprusik578 2020-05-14 18:20:23
8 and 9 only APPEAR ONCE
DPatrick 2020-05-14 18:20:32
Hmmm...some of the cycles take 4 steps to go between $1$s, and some only take 3 steps.
DPatrick 2020-05-14 18:20:39
And it's the $4$-step cycles that produce $8$s and $9$s.
DPatrick 2020-05-14 18:20:53
Do you see how to put all of these pieces together?
DPatrick 2020-05-14 18:21:03
There's one more bit of data we haven't used yet. (This is a great problem-solving strategy: if you're stuck, look back and see what you haven't used yet.)
sixoneeight 2020-05-14 18:21:13
Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 ones.
superagh 2020-05-14 18:21:13
use the 302 digits information
Maths4J 2020-05-14 18:21:13
We know the number of cycles too, 302
sixoneeight 2020-05-14 18:21:20
Whenever there is a new one, there must be a digit change, so we can conclude that there are 302 of them
Ninja11 2020-05-14 18:21:26
How does 2^1000 with 302 digits help?
Doudou_Chen 2020-05-14 18:21:28
is it the 302 digits part?
pi_is_3.14 2020-05-14 18:21:31
$2^100$ has 302 digits
DPatrick 2020-05-14 18:21:37
We haven't yet used the given fact that $2^{1000}$ has $302$ digits.
DPatrick 2020-05-14 18:21:48
They must have told us that for a reason
Maths4J 2020-05-14 18:22:00
actually, it says 2^1000 is a 302 digit number starting with 1, it starts the 302nd cycle
onedance 2020-05-14 18:22:04
2^1000 has 302 digits?
fasterthanlight 2020-05-14 18:22:16
so there are 301 cycles
DPatrick 2020-05-14 18:22:24
Right! From our initial $1$, we have to go through $301$ cycles to get to the final $1$. (We have to add $301$ digits to get from the $1$-digit $2^0$ to the $302$-digit $2^{1000}.)$
DPatrick 2020-05-14 18:22:47
And since we're going from $2^0$ to $2^{1000}$, it'll take us $1000$ steps to do so.
DPatrick 2020-05-14 18:22:58
So if we use $301$ cycles to take $1000$ steps, how many of those cycles are $3$-steps and how many are $4$-steps?
DPatrick 2020-05-14 18:24:01
Actually, we don't really care how many are $3$-step, since what we care about (the $9$s) only appear in the $4$-steps.
palindrome868 2020-05-14 18:24:13
97 4-steps
97 4-steps
Math5K 2020-05-14 18:24:13
97 4-step
yayy 2020-05-14 18:24:13
97 4 step
Leonard_my_dude 2020-05-14 18:24:13
97 4 steps
bobjoebilly 2020-05-14 18:24:13
97 4-step
Champion1234 2020-05-14 18:24:13
97 4 steps
green_alligator 2020-05-14 18:24:13
97 4 steps
sixoneeight 2020-05-14 18:24:18
x+y=301, 3x+4y=1000
Mathematician1010 2020-05-14 18:24:18
$3x+4y=1000$

$x+y=301$
jai123 2020-05-14 18:24:18
x+y=301 and 3x+4y=1000
DPatrick 2020-05-14 18:24:36
You could certainly set up a system of equations to solve this.
DPatrick 2020-05-14 18:24:53
Here's how I think to think about it: "If every cycle took 3 steps, we'd only take 903 steps. We need 97 extra steps to make it to 1000."
DPatrick 2020-05-14 18:25:05
That is, since $3 \cdot 301 = 903$, we need $1000 - 903 = 97$ "extra" steps to make it to $1000$.
DPatrick 2020-05-14 18:25:13
So $97$ of the cycles are $4$-step cycles.
DPatrick 2020-05-14 18:25:15
And now what?
52 have 8's so 45 have 9's!
gs_2006 2020-05-14 18:25:35
and now you subtract 52 from 97
Jomo 2020-05-14 18:25:35
So 97-52=45?
fasterthanlight 2020-05-14 18:25:35
97 4-steps, 52 are 1-2-4-8, so 9 comes out 45 times
fasterthanlight 2020-05-14 18:25:35
52 are 1-2-4-8
mfro24 2020-05-14 18:25:35
We subtract 52!
E1CAarush 2020-05-14 18:25:35
97-52
Leonard_my_dude 2020-05-14 18:25:35
97 - 52 = 45
CHIPPER33 2020-05-14 18:25:35
subract 52
kred9 2020-05-14 18:25:35
97-52=45 yay
millburn2006 2020-05-14 18:25:35
97-52=45
The_Better_Samuel 2020-05-14 18:25:35
97-52=45 so 45 times 9 appears
PShucks 2020-05-14 18:25:35
subtract 45 out
EpicGirl 2020-05-14 18:25:35
45 p's
qsheng 2020-05-14 18:25:35
97-52=45
DPatrick 2020-05-14 18:25:42
Aha, we're at the finish line!
DPatrick 2020-05-14 18:25:46
Each of the $97$ $4$-step cycles contains either an $8$ or a $9$ (but not both).
DPatrick 2020-05-14 18:26:01
But we're told that $52$ of them contain an $8.$ (Yet another bit of data from the problem that we haven't used yet!)
DPatrick 2020-05-14 18:26:13
Therefore, the other $97 - 52 = \boxed{45}$ of them contain a $9$, and we're done!
DPatrick 2020-05-14 18:26:36
I really liked this problem: all the clues were there, but you had to work pretty hard to put them together to solve the mystery.
DPatrick 2020-05-14 18:27:07
We're going to finish up tonight by looking at the final two Team problems.
DPatrick 2020-05-14 18:27:18
I know we've been here a while, so if you have to go, that's OK!
DPatrick 2020-05-14 18:27:29
We'll post a transcript of the entire session on the website when we're done.
smartguy888 2020-05-14 18:27:44
"rrusczyk:We're not sure! Maybe 90 minutes or so." Reality: 3 hours XD
DPatrick 2020-05-14 18:27:59
Yeah, we were a little off. We've never done a Math Jam like this before!
rrusczyk 2020-05-14 18:28:02
Harvey has been mocking me for an hour and a half.
rrusczyk 2020-05-14 18:28:10
Please don't give him any more ammunition.
rrusczyk 2020-05-14 18:28:22
Meanwhile, the Other Guy told me he would have finished 2 hours ago.
DPatrick 2020-05-14 18:28:33
This has blown away our record for number of people at a Math Jam. (I believe the old record was a little over 800. We had over 1300 today.)
DPatrick 2020-05-14 18:28:53
So let's look at Team #9 next:
DPatrick 2020-05-14 18:28:58
Team #9: A cheerleading squad has $14$ cheerleaders, each a different height. How many ways are there for the cheerleaders to line up for a photo in two rows with seven people each, so that each cheerleader in the back row is taller than the one immediately in front of them, and so that the heights of the cheerleaders in the back row descend from the middle to each side?
Quaoar 2020-05-14 18:29:27
Ummmm...
CT17 2020-05-14 18:29:27
Wow
DPatrick 2020-05-14 18:29:42
Those were my first thoughts too.
SharonW 2020-05-14 18:29:53
sdattilo2002 2020-05-14 18:29:53
that's complicated
mewto 2020-05-14 18:29:53
that looks complicated
DPatrick 2020-05-14 18:30:05
We have a lot of information to try to organize.
winterrain01 2020-05-14 18:30:18
Just assign the cheerleaders heights of 1, 2, 3, ..., 14!
DPatrick 2020-05-14 18:30:35
Great idea! So that they're easier to talk about, let's number the cheerleaders from $1$ to $14$ in order of height, with $1$ being the shortest and $14$ being the tallest.
mahaler 2020-05-14 18:31:10
I think first we should draw it out
TMSmathcounts737 2020-05-14 18:31:10
Let's start to draw a picture
DPatrick 2020-05-14 18:31:27
I wholeheartedly agree.
DPatrick 2020-05-14 18:31:32
A great problem solving strategy is to draw a picture.
DPatrick 2020-05-14 18:31:36
We can convert the text of the problem statement into an easier-to-use diagram:
DPatrick 2020-05-14 18:31:39
DPatrick 2020-05-14 18:31:53
The $<$ signs are the usual greater-than signs: they indicate where a cheerleader must have a larger number (that is, be taller) than the cheerleader standing next to him or her.
DPatrick 2020-05-14 18:32:02
(Or less-than signs, I guess!)
DPatrick 2020-05-14 18:32:10
So we need to count the number of ways we can fill the boxes with the numbers $1$ to $14$, preserving all the inequalities shown.
DPatrick 2020-05-14 18:32:21
Notice anything quick?
Rsar12 2020-05-14 18:32:44
14 must be in the back middle
Streaks123 2020-05-14 18:32:44
14 must go in center back.
PrinceRoyale 2020-05-14 18:32:44
tallest(14) has to take the back middle spot, because he is the tallest
MR_67 2020-05-14 18:32:44
The middle back person is the tallest
albertedwin 2020-05-14 18:32:44
c#14 has to be in the middle of the back row.
fasterthanlight 2020-05-14 18:32:44
14 is in the middle position of the back row.
maxben 2020-05-14 18:32:44
14 should be in te middle of the second row.
$14$ has to be in the back center
nathanqiu 2020-05-14 18:32:44
so middle back is 14
EulerRocks2.718 2020-05-14 18:32:44
14 must go in the top middle
14 HAS to be in the middle of the back row
bobjoebilly 2020-05-14 18:32:44
The one in the middle of the back row needs to be #14.
austinchen2005 2020-05-14 18:32:44
14 is in the middle!
DPatrick 2020-05-14 18:32:55
Yeah! We notice that the $14$ must be in the center of the back row, as this is the only box in the diagram that is not less than some other box.
DPatrick 2020-05-14 18:33:03
DPatrick 2020-05-14 18:33:15
Good start! Always good to get the easiest stuff out of the way.
DPatrick 2020-05-14 18:33:17
Now what?
jupiter314 2020-05-14 18:34:07
there are now 3 ways for 13. either side of 14 or in front of 14
edjar 2020-05-14 18:34:07
3 choices for 13
MathJams 2020-05-14 18:34:07
13 could go in 3 places
cat_maniac_8 2020-05-14 18:34:07
13 has to be surrounding it
bacond 2020-05-14 18:34:07
The 13 has to be on the side of the 14
kkomma10 2020-05-14 18:34:07
the next one is 13
yayy 2020-05-14 18:34:07
13 is next to it
MAthen07 2020-05-14 18:34:07
13 has to be next to 14.
DPatrick 2020-05-14 18:34:34
I agree: given what's left, the $13$ has to either be in the back row next to $14$, or could be directly in front of $14$.
DPatrick 2020-05-14 18:34:52
Every other box is shorter than another box that's still empty.
DPatrick 2020-05-14 18:35:10
This feels like that this sort of "constructive counting" is starting to get ugly.
DPatrick 2020-05-14 18:35:36
In counting problems, I often like to step back and take a look at the big picture before diving into too much casework.
DPatrick 2020-05-14 18:35:54
Do you have any "big picture" observations about our diagram?
DPatrick 2020-05-14 18:35:58
harmonyguan 2020-05-14 18:36:18
symmetrical?
pi_is_3.14 2020-05-14 18:36:23
symmetry
Eng123 2020-05-14 18:36:23
Symmetry.
DPatrick 2020-05-14 18:36:41
Indeed: the left side and the right side look like mirror images of the same picture.
PrinceRoyale 2020-05-14 18:37:03
you can cut it in half and solve one side only. but remember to multiply at the end
DPatrick 2020-05-14 18:37:23
Aha -- it looks like the left side and the right side are essentially the same.
sixoneeight 2020-05-14 18:37:28
First, you have to choose the one in front of 14
DPatrick 2020-05-14 18:37:40
True, let's not forget about that too!
DPatrick 2020-05-14 18:37:57
So it looks like we've observed that the diagram splits into three pieces from here:
DPatrick 2020-05-14 18:38:01
DPatrick 2020-05-14 18:38:17
That is, the remaining cheerleaders can be divided into three groups: $6$ to go in the three columns on the left side shaded in blue, $6$ to go in the three columns on the right side shaded in green, and the thirteenth to go in the center of the front row shaded in pink.
casp 2020-05-14 18:38:40
there are 13 ways
yangi26 2020-05-14 18:38:40
you have 13 choices for the one in front of 14
nathanqiu 2020-05-14 18:38:40
13 choices for pink
jupiter314 2020-05-14 18:38:40
there are 13 ways to choose the person in front.
DPatrick 2020-05-14 18:39:00
Good, let's do that next: we have $13$ choices for who goes in the pink spot in front of $14$. Anybody can go there!
DPatrick 2020-05-14 18:39:10
So now we have twelve of them left.
DPatrick 2020-05-14 18:39:21
How do we split them into two groups of six, to go on either side?
sixoneeight 2020-05-14 18:40:00
12 choose 6
krishgarg 2020-05-14 18:40:00
12 choose 6
kred9 2020-05-14 18:40:00
12C6?
kdraganov 2020-05-14 18:40:00
$\dbinom{12}{6}$
Eng123 2020-05-14 18:40:00
12 choose 6.
darthtator012 2020-05-14 18:40:00
12 choose 6
Rsar12 2020-05-14 18:40:00
12 choose 6?
mohanty 2020-05-14 18:40:00
we pick any 6 out of 12
mfro24 2020-05-14 18:40:00
We have 12c6 ways to do it!
Alculator11 2020-05-14 18:40:00
$\binom{12}{6}$
MR_67 2020-05-14 18:40:00
12C6
DPatrick 2020-05-14 18:40:17
Right: this is a job for combinations!
DPatrick 2020-05-14 18:40:45
We can choose $6$ of the $12$ to go on the blue side in $\dbinom{12}{6}$ ways.
DPatrick 2020-05-14 18:40:57
Then the remaining $6$ will go on the green side.
fasterthanlight 2020-05-14 18:41:11
924 ways
DarrenHuang888 2020-05-14 18:41:11
924 is 12 choose 6
mmjguitar 2020-05-14 18:41:11
924
Beastboss 2020-05-14 18:41:11
924 ways
manokid 2020-05-14 18:41:11
924
DPatrick 2020-05-14 18:41:22
Right: good thing that calculator is still handy!
DPatrick 2020-05-14 18:41:28
$\dbinom{12}{6} = 924$
DPatrick 2020-05-14 18:41:51
So to recap the choices so far: we had $13$ choices for the pink, then $924$ choices for the group of $6$ to be blue, and the rest are green.
then you order them!
DPatrick 2020-05-14 18:42:23
Right, there's still more do to! For each of the groups of $6$ to one side, we have to count the number of ways they can be arranged.
DPatrick 2020-05-14 18:42:49
Since only the order of their heights matter, let's relabel them $A$ (shortest) through $F$ (tallest), just so it's easier to keep track of them.
DPatrick 2020-05-14 18:43:05
Oops
DPatrick 2020-05-14 18:43:08
So for the left side, we need to count the ways to place $A$ through $F$ in the diagram below, such that all of the inequality signs work:
DPatrick 2020-05-14 18:43:12
cj13609517288 2020-05-14 18:43:36
Hey, we know where to put the greatest number in the 6-group!
dolphin7 2020-05-14 18:43:36
F must go next to 14
Alculator11 2020-05-14 18:43:36
F goes in the top right
Awesome3.14 2020-05-14 18:43:36
F goes on the top right
Quaoar 2020-05-14 18:43:36
Tallest in the top right
cj13609517288 2020-05-14 18:43:36
F goes in top right
vtlev 2020-05-14 18:43:36
F is in op right
E1CAarush 2020-05-14 18:43:36
The top right is the greatest
bobjoebilly 2020-05-14 18:43:36
F must be in the upper right hand corner.
FearlessTaurus 2020-05-14 18:43:36
F goes in the top right
Speedstorm 2020-05-14 18:43:36
F is in the top right
DPatrick 2020-05-14 18:43:46
Good: we see that $F$ must go in the upper-right box, as every other box is less than some other box.
DPatrick 2020-05-14 18:43:49
cat_maniac_8 2020-05-14 18:44:21
E is either bellow are to the left of F
Kruxe 2020-05-14 18:44:21
then E could go in the one to the left or the one underneath
melonlord 2020-05-14 18:44:21
E is in the top middle or bottom right
sri1priya 2020-05-14 18:44:21
E has 2 choices?
sigma_notation 2020-05-14 18:44:21
E must go next to F
peace09 2020-05-14 18:44:21
E is next to F or below F
DPatrick 2020-05-14 18:44:29
We see that $E$ must either go in the lower-right box or the upper-center box.
DPatrick 2020-05-14 18:44:44
I don't think there's anything much clever we can do except to count the two cases.
DPatrick 2020-05-14 18:44:54
Fortunately the diagram is small, so hopefully it'll not be too bad.
DPatrick 2020-05-14 18:45:00
Case 1: we place $E$ in the lower-right box:
DPatrick 2020-05-14 18:45:03
DPatrick 2020-05-14 18:45:14
How do we finish with A-D?
Alculator11 2020-05-14 18:45:40
Now we have 1 choice for D
nathanqiu 2020-05-14 18:45:40
D is left of F
Gentoo 2020-05-14 18:45:40
Then D has to go next to F
sdattilo2002 2020-05-14 18:45:40
d goes next to f
D must be right next to F
PShucks 2020-05-14 18:45:40
d must be top center
DPatrick 2020-05-14 18:45:45
Then $D$ must go in the upper-center box.
DPatrick 2020-05-14 18:45:48
DPatrick 2020-05-14 18:46:11
How do we finish this case?
melonlord 2020-05-14 18:46:23
C is in the top left or bottom middle
jupiter314 2020-05-14 18:46:23
There are 2 choices for C
cat_maniac_8 2020-05-14 18:46:23
C goes either below or to the left of D
MR_67 2020-05-14 18:46:23
C goes either middle bottom or top left
Alculator11 2020-05-14 18:46:30
Now 2 choices for C
manokid 2020-05-14 18:46:30
Then c can go below or next to d
Unicorn78 2020-05-14 18:46:30
C has two choices
BakedPotato66 2020-05-14 18:46:30
C can either go to the left or in front of D
DPatrick 2020-05-14 18:46:35
Right.
DPatrick 2020-05-14 18:46:53
We could put $C$ in the bottom-middle, and then $A$ and $B$ are forced:
DPatrick 2020-05-14 18:46:57
DPatrick 2020-05-14 18:47:08
Or else $C$ goes in the upper-left, in which case $A$ and $B$ can fill the remaining two spots in either order:
DPatrick 2020-05-14 18:47:12
DPatrick 2020-05-14 18:47:16
DPatrick 2020-05-14 18:47:38
So that's $3$ ways to place the group of six in this case.
DarrenHuang888 2020-05-14 18:47:46
we could have E next to F too
DPatrick 2020-05-14 18:47:55
Right, the other case was that $E$ ends up next to $F$:
DPatrick 2020-05-14 18:47:59
DPatrick 2020-05-14 18:48:11
How do we finish counting this case from here?
DPatrick 2020-05-14 18:48:48
There are lots of cases for $D$, but is there a simpler way to count this case?
cj13609517288 2020-05-14 18:49:02
24(total)/2(only half work for the left)=12
DPatrick 2020-05-14 18:49:17
That's a great observation!
DPatrick 2020-05-14 18:49:33
There are $4! = 24$ ways to place $A$ through $D$ in any fashion.
DPatrick 2020-05-14 18:49:56
But exactly half of them work in the first column, with a taller person in back. By symmetry, the other half illegally have a taller person in front.
DPatrick 2020-05-14 18:50:03
So there are $4!/2 = 12$ ways to finish this case.
DPatrick 2020-05-14 18:50:16
Adding the cases for the group of $6$ on the left, how many ways to arrange them?
TheEpicCarrot7 2020-05-14 18:50:33
3+12=15
Jomo 2020-05-14 18:50:33
15
edjar 2020-05-14 18:50:33
15
dolphin7 2020-05-14 18:50:33
15
winner320 2020-05-14 18:50:33
15
Zhaom 2020-05-14 18:50:33
15
DPatrick 2020-05-14 18:50:37
Adding our two cases gives us $3+12 = 15$ ways to arrange the cheerleaders on the left side.
Quaoar 2020-05-14 18:51:02
same for right side
edjar 2020-05-14 18:51:02
And 15 more on the right
karthic7073 2020-05-14 18:51:02
And it would be the same for the other side!
Speedstorm 2020-05-14 18:51:02
so there's 15 ways to arrange them on the right side as well
DPatrick 2020-05-14 18:51:12
Right, don't forget! Since the right side is exactly the same (it's just a mirror image of the left side), there are also $15$ ways to arrange the cheerleaders on the left side.
mfro24 2020-05-14 18:51:23
So we have a total of $13\cdot\binom{12}{6}\cdot(15^2)$, right?
DPatrick 2020-05-14 18:51:44
Exactly. If we go back and look at all the choices we had along the way:
DPatrick 2020-05-14 18:51:51
We had $13$ choices for the pink person,
DPatrick 2020-05-14 18:52:02
We had $\dbinom{12}{6}$ choices for the split into two groups of six.
DPatrick 2020-05-14 18:52:09
And we had $15$ choices for each of the two groups.
DPatrick 2020-05-14 18:52:23
So the total count is $13 \cdot \dbinom{12}{6} \cdot 15^2$
fasterthanlight 2020-05-14 18:52:40
2,702,700 ways!
so there are 2702700 number of cases total
fasterthanlight 2020-05-14 18:52:40
2,702,700 ways to arrange the cheerleaders
mfro24 2020-05-14 18:52:40
Which gives us a grand total of $2,702,700$!
FSMS-BL 2020-05-14 18:52:40
now we just put it into a calculator
MR_67 2020-05-14 18:52:40
rneelam 2020-05-14 18:52:40
$2702700$ is the answer!
MR_67 2020-05-14 18:52:40
2702700
Jomo 2020-05-14 18:52:40
$$13 \times 924 \times 15^2 = 2,702,700‬$$
manokid 2020-05-14 18:52:40
2702700
DPatrick 2020-05-14 18:52:46
Therefore, our final count is $12012 \cdot 225 = \boxed{2702700}$ (ways).
winterrain01 2020-05-14 18:52:58
A nice number
Awesome3.14 2020-05-14 18:52:58
wow that is a lot!
DPatrick 2020-05-14 18:53:22
OK, back to rrusczyk for Team #10!
sdattilo2002 2020-05-14 18:53:39
last one!
is this the last one
rrusczyk 2020-05-14 18:53:44
Last problem for today!
rrusczyk 2020-05-14 18:53:48
Thanks for staying with us!
Alculator11 2020-05-14 18:53:53
@rrusczyk Why isn't Harvey in this Math Jam?
rrusczyk 2020-05-14 18:53:59
But he is! Don't you see him?
Jerry_Guo 2020-05-14 18:54:41
does harvey exist?
rrusczyk 2020-05-14 18:54:45
He sure thinks he does.
ohjoshuaoh 2020-05-14 18:54:48
harvey is the best, no offense Mr. Rusczyk
rrusczyk 2020-05-14 18:54:57
None taken. Harvey is pretty awesome.
rrusczyk 2020-05-14 18:55:01
Now, time for the last problem.
rrusczyk 2020-05-14 18:55:10
Thanks for staying with us for so long. Nearly 4 hours.
rrusczyk 2020-05-14 18:55:15
And still over 700 people here!
rrusczyk 2020-05-14 18:55:27
Team #10: If six numbers are chosen at random, with replacement, from the set of integers from $1$ to $900,$ inclusive, what is the probability that the product of these six integers leaves a remainder of $4$ when divided by $30?$ Express your answer as a percent to the nearest thousandth.
rrusczyk 2020-05-14 18:55:32
Let's start off by calling our product $P$ so we can talk about it.
rrusczyk 2020-05-14 18:55:46
What does "$P$ leaves a remainder of $4$ when divided by $30$" tell us about $P$ that might be helpful?
Jomo 2020-05-14 18:56:01
It must be divisible by 2.
cj13609517288 2020-05-14 18:56:01
P is even.
Speedstorm 2020-05-14 18:56:01
It's even
Derpy123 2020-05-14 18:56:01
p is even
Eng123 2020-05-14 18:56:02
Even.
rrusczyk 2020-05-14 18:56:07
Well, for starters, $P$ is clearly even. "$P$ leaves a remainder of $4$ when divided by $30$" means we can write $P$ as $30k+4$ for some integer $k.$
rrusczyk 2020-05-14 18:56:08
So, $P = 30k+4 = 2(15k+2),$ which means $P$ is even.
rocketsri 2020-05-14 18:56:19
P is 4 mod 30
Alculator11 2020-05-14 18:56:19
$P\equiv 4\mod 30$
Notaval 2020-05-14 18:56:19
4 mod 30
rrusczyk 2020-05-14 18:56:57
A lot of you are talking about mods. That's great! Let's see how far we can get without ever using modular arithmetic -- a lot of MATHCOUNTS students haven't learned modular arithmetic yet. (If you haven't, go check it out.)
rrusczyk 2020-05-14 18:57:04
So, $P$ is even.
rrusczyk 2020-05-14 18:57:06
What else?
thanosaops 2020-05-14 18:57:25
P is not divisible by either 3 or 5
Notaval 2020-05-14 18:57:25
not divisible by 3
MR_67 2020-05-14 18:57:25
P is not both a multiple of 3 and 5
thanosaops 2020-05-14 18:57:25
P is not divisible by 3 or 5
Jomo 2020-05-14 18:57:25
P is not divisible by 3.
MathJams 2020-05-14 18:57:40
P is not a multiple of 3 or 5
rrusczyk 2020-05-14 18:57:54
It's not a multiple of 3 or 5. Could it be any odd that's not a multiple of 3 or 5?
kylewu0715 2020-05-14 18:58:11
seven
NO IT HAS TO BE EVEN!
rrusczyk 2020-05-14 18:58:30
Oops, I mean could it be any even that's not a multiple of 3 or 5?
BakedPotato66 2020-05-14 18:58:51
8
cj13609517288 2020-05-14 18:58:54
it can't, like 8 surpasses
sigma_notation 2020-05-14 18:58:57
8 doesn't work, so no
rrusczyk 2020-05-14 18:59:01
Can't be 8 either.
rrusczyk 2020-05-14 18:59:11
Why not?
Lux1 2020-05-14 18:59:18
it has to leave a remainder of 4 when divided by 30
Dalar25 2020-05-14 18:59:27
it has to have a remainder of 4
winner320 2020-05-14 18:59:27
it has to have remainder 4
rrusczyk 2020-05-14 18:59:28
Great.
rrusczyk 2020-05-14 18:59:40
So, we've already seen what must happen if we divide by 2.
rrusczyk 2020-05-14 18:59:53
We've seen that we can't get an integer if we divide by 3.
rrusczyk 2020-05-14 19:00:05
Can we figure anything else out about dividing by 3?
yayatheduck 2020-05-14 19:00:28
remainder 1
MathJams 2020-05-14 19:00:28
leaves a remainder of 1
sixoneeight 2020-05-14 19:00:28
remainder 1
Rsar12 2020-05-14 19:00:28
leaves a remainder of 1
MAthen07 2020-05-14 19:00:28
Remainder 1
Alculator11 2020-05-14 19:00:28
We get a remainder of 1
mfro24 2020-05-14 19:00:28
It has a remainder of 1!
maxben 2020-05-14 19:00:28
remainder is 1
edjar 2020-05-14 19:00:28
must be 1 more than multiple of 3
rrusczyk 2020-05-14 19:00:45
Exactly. For those of you new to "modular arithmetic", we say:
Alculator11 2020-05-14 19:00:48
It's 1 mod 3
rrusczyk 2020-05-14 19:01:00
That's all "1 mod 3" means -- 1 more than a multiple or 3.
rrusczyk 2020-05-14 19:01:04
$P$ must be one more than a multiple of $3,$ since $P=30k+4=30k+3+1=3(10k+1)+1.$
rrusczyk 2020-05-14 19:01:13
All right, and what happens if we divide by 5?
MathJams 2020-05-14 19:01:37
remainder of 4
remain 4
rocketsri 2020-05-14 19:01:37
Remainder 4
Math5K 2020-05-14 19:01:37
remainder 4
Jomo 2020-05-14 19:01:37
4 more.
MAthen07 2020-05-14 19:01:37
Remainder 4
TheEpicCarrot7 2020-05-14 19:01:37
4 as a remainder
remainder 4
yayatheduck 2020-05-14 19:01:37
remainder 4
DarrenHuang888 2020-05-14 19:01:37
remainder of 4
manokid 2020-05-14 19:01:37
Remainder of 4
millburn2006 2020-05-14 19:01:37
remainder of 4
rrusczyk 2020-05-14 19:01:40
$P$ must be $4$ more than a multiple of $5,$ since $P = 30k+4 = 5(6k)+4.$
rrusczyk 2020-05-14 19:01:49
Another way to say this:
sdattilo2002 2020-05-14 19:01:52
it's 4 mod 5
jupiter314 2020-05-14 19:01:52
it's 4 mod 5
rrusczyk 2020-05-14 19:01:55
Now...
rrusczyk 2020-05-14 19:01:57
Why are we focusing on $2, 3,$ and $5$ here?
Math5K 2020-05-14 19:02:31
factors of 30
factors of 30
sixoneeight 2020-05-14 19:02:31
factors of 30
MR_67 2020-05-14 19:02:31
they are the factors of 30\
aidni47 2020-05-14 19:02:31
factors of 30
fasterthanlight 2020-05-14 19:02:31
they are factors of 30
TheEpicCarrot7 2020-05-14 19:02:31
because they are the prime factors of 30
IAM_TSSXII 2020-05-14 19:02:31
prime factorization of 30
fasterthanlight 2020-05-14 19:02:31
prime factors of 30
Gentoo 2020-05-14 19:02:31
Because those are the prime factors of 30
Unicorn78 2020-05-14 19:02:31
All prime factors of 30
SharonW 2020-05-14 19:02:31
they are prime factors of 30
ChrisalonaLiverspur 2020-05-14 19:02:45
because they are the prime factors of 30.
rrusczyk 2020-05-14 19:02:49
Ah, yes, they're the prime factors of $30.$
rrusczyk 2020-05-14 19:02:50
OK, so now we know that $P$ is even, $1$ more than a multiple of $3,$ and $4$ more than a multiple of $5.$ We can write this as

\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$
rrusczyk 2020-05-14 19:02:58
But... Does this go the other way? That is, must any $P$ that satisfies all three of these give a remainder of $4$ when divided by $30?$
edjar 2020-05-14 19:03:12
Yes
kred9 2020-05-14 19:03:12
yes
rrusczyk 2020-05-14 19:03:17
Let's see! What do the first two give us? ($P=2a$ and $P=3b+1.$)
TThB0501 2020-05-14 19:03:35
b is odd
FearlessTaurus 2020-05-14 19:03:35
b is odd
MR_67 2020-05-14 19:03:35
b is odd
Jomo 2020-05-14 19:03:35
b must be odd.
fasterthanlight 2020-05-14 19:03:43
b is odd
arnav21 2020-05-14 19:03:43
b is odd
bacond 2020-05-14 19:03:44
b is odd
akpi 2020-05-14 19:03:44
b is odd
rrusczyk 2020-05-14 19:03:50
Since $P$ must be even, we know that $b$ must be odd. So, we have $b=2d+1$ for some integer $d,$ and

$P=3b+1=3(2d+1)+1=6d+4.$
rrusczyk 2020-05-14 19:04:23
Notice that "$b$ is odd means $b=2d+1$ for some integer $d$" step -- these sorts of observations can be really helpful in number theory problems!
rrusczyk 2020-05-14 19:04:29
And what happens when we combine this with $P = 5c+4?$
rrusczyk 2020-05-14 19:04:49
We have $P = 6d+4$ and $P=5c+4,$ and $d$ and $c$ are integers...
SharonW 2020-05-14 19:04:59
5c=6d
sarahAops2020 2020-05-14 19:04:59
6d=5c
edjar 2020-05-14 19:04:59
5c = 6 d
FSMS-BL 2020-05-14 19:04:59
6d = 5c
E1CAarush 2020-05-14 19:04:59
6d = 5c
HarleyMathCounts 2020-05-14 19:04:59
5c=6d
peace09 2020-05-14 19:05:13
6d=5c
rrusczyk 2020-05-14 19:05:14
Combining $P=6d+4$ and $P=5c+4$ gives $6d+4=5c+4,$ so $6d=5c.$
rrusczyk 2020-05-14 19:05:27
Again, $c$ and $d$ are integers, so what do we know now?
c is a multiple of 6
Math5K 2020-05-14 19:05:44
and d is multiple of 5
d is a multiple of 5
MR_67 2020-05-14 19:05:44
c is a mult of 6
sixoneeight 2020-05-14 19:05:44
d is a multipe of 5
nathanqiu 2020-05-14 19:05:44
d is divisible by 5
wlm2 2020-05-14 19:05:44
d is a multiple of 5 and c is a multiple of 6
jupiter314 2020-05-14 19:05:46
d is a multiple of 5, c is a multiple of 6
rrusczyk 2020-05-14 19:06:05
$6d=5c$ tells us that $d$ is a multiple of $5,$ and $c$ is a multiple of $6.$
rrusczyk 2020-05-14 19:06:40
Since $c$ is a multiple of $6,$ we have $c=6f$ for some integer $f.$ What does that give us?
TheEpicCarrot7 2020-05-14 19:07:10
P=30f+4
FSMS-BL 2020-05-14 19:07:10
30f+4
Alculator11 2020-05-14 19:07:10
P=30f+4
bobjoebilly 2020-05-14 19:07:22
P = 30f+4
rrusczyk 2020-05-14 19:07:23
Letting $c = 6f,$ we now have

$P=5c+4 = 5(6f)+4=30f+4.$
nathanqiu 2020-05-14 19:07:33
Jomo 2020-05-14 19:07:33
which is what we already knew...
peace09 2020-05-14 19:07:33
oof circular reasoning
rrusczyk 2020-05-14 19:07:57
Not quite! When we started, we said that $P=30k+4$ means that the following is true:
rrusczyk 2020-05-14 19:08:08
\begin{align*}

P &=2a,\\

P&=3b+1,\\

P&=5c + 4,

\end{align*}

for some integers $a,b,c.$
rrusczyk 2020-05-14 19:08:15
We JUST WENT THE OTHER WAY!
sixoneeight 2020-05-14 19:08:30
we had to prove that it works the other way
TheEpicCarrot7 2020-05-14 19:08:30
We proved the convers
rrusczyk 2020-05-14 19:08:45
We started with those three equations and showed that the result is that $P=30f+4$ for some integer $f.$
arnav21 2020-05-14 19:08:58
Wow
Alculator11 2020-05-14 19:08:58
So we've shown that it's necessary and sufficient
MathWiz20 2020-05-14 19:08:58
We proved it
akpi 2020-05-14 19:08:58
we proved that it works the other way
peace09 2020-05-14 19:08:58
we proved the converse... yea
PShucks 2020-05-14 19:08:59
rrusczyk 2020-05-14 19:09:13
Great question. Why do we like these three new equations?
rrusczyk 2020-05-14 19:09:23
(What we just worked through here is an example of the Chinese Remainder Theorem, which is really powerful. You might want to check out the Chinese Remainder Theorem later.)
cj13609517288 2020-05-14 19:09:37
Since they are easier to use
rocketsri 2020-05-14 19:09:37
They are simpler
rrusczyk 2020-05-14 19:09:47
Exactly, we've broken our problem down to three smaller problems.
rrusczyk 2020-05-14 19:09:59
Let's start with the first: what is the probability that our product ends up even?
Trollyjones 2020-05-14 19:10:30
1/2
sdattilo2002 2020-05-14 19:10:30
1/2
Jomo 2020-05-14 19:10:30
1/2
drakewilson 2020-05-14 19:10:30
1/2
Lux1 2020-05-14 19:10:30
1/2
50%
manokid 2020-05-14 19:10:30
1/2
mmjguitar 2020-05-14 19:10:30
1/2
awesomebooks 2020-05-14 19:10:30
1/2
MathJams 2020-05-14 19:10:30
1/2
palindrome868 2020-05-14 19:10:30
1/2
felicitas 2020-05-14 19:10:30
1/2
rrusczyk 2020-05-14 19:10:51
That's the probability a single number we choose is even.
rrusczyk 2020-05-14 19:11:01
We need the product of all six integers to be even.
rrusczyk 2020-05-14 19:11:15
How do we figure that out?
yayy 2020-05-14 19:11:36
1 - 1/2^6
maxben 2020-05-14 19:11:36
1-(1/2)^6
NinjaMango 2020-05-14 19:11:36
1 - (1/2)^6 = 63/64
Jerry_Guo 2020-05-14 19:11:36
63/64
Streaks123 2020-05-14 19:11:36
63/64
1-1/64=63/64
yangi26 2020-05-14 19:11:39
probability they are ALL odd
skyguy88 2020-05-14 19:11:40
Complementary counting (odds of it being odd) gives 1 - 1/64 = 63/64
logz 2020-05-14 19:11:43
$1-\frac{1}{2^6}$
melonlord 2020-05-14 19:11:47
find the probability of the product being odd
rrusczyk 2020-05-14 19:11:50
In order to be even, we can't have all $6$ chosen numbers be odd.
rrusczyk 2020-05-14 19:11:52
All $6$ are odd with probability $\left(\frac12\right)^6=\frac{1}{64},$ so the probability the product is even is

$1-\frac{1}{64} = \frac{63}{64}.$
rrusczyk 2020-05-14 19:12:20
Next in the three things we know about our product -- it must have a remainder of $1$ when divided by $3.$
rrusczyk 2020-05-14 19:12:54
What must be true about the numbers we choose?
Alculator11 2020-05-14 19:12:58
So it can't be divisible by 3
MAthen07 2020-05-14 19:13:03
There can be no number chosen that is divisible by 3.
TheEpicCarrot7 2020-05-14 19:13:15
None of them can be multiples of 3
melonlord 2020-05-14 19:13:15
none of them are divisible by 3
aidni47 2020-05-14 19:13:15
none of them can be a mult of 3
rrusczyk 2020-05-14 19:13:20
Ah, that's a good start!
rrusczyk 2020-05-14 19:13:33
First up, none of the chosen numbers can be a multiple of $3.$ What's the probability that this happens?
fasterthanlight 2020-05-14 19:14:19
64/729
melonlord 2020-05-14 19:14:19
(2/3)^6
mmjguitar 2020-05-14 19:14:19
(2/3)^6
ElNoraa 2020-05-14 19:14:19
64/729
Jomo 2020-05-14 19:14:19
$(2/3)^6 = 64/729$
rrusczyk 2020-05-14 19:15:41
Each number has a $\frac23$ chance of not being a multiple of $3,$ so the probability that the product is NOT a multiple of $3$ is

$\left(\frac23\right)^6 = \frac{64}{729}.$
rrusczyk 2020-05-14 19:16:08
Um, but then our product might be either $1$ or $2$ more than a multiple of $3.$ Now what?
johnnyboy1113 2020-05-14 19:16:24
reduce by 50%
szheng0312 2020-05-14 19:16:24
divide by 2
E1CAarush 2020-05-14 19:16:24
Divide by 2
karthic7073 2020-05-14 19:16:25
divide by 2!
BakedPotato66 2020-05-14 19:16:27
divide by two- two cases are symmetrical
Math4Life2020 2020-05-14 19:16:29
Times 1/2
cj13609517288 2020-05-14 19:16:36
are you sure.
rrusczyk 2020-05-14 19:16:45
Good question -- why can we just take half?
palindrome868 2020-05-14 19:17:14
half of them give us 1
MR_67 2020-05-14 19:17:14
half is 1 more, half is 2 more than a mult of 3
Eagle21 2020-05-14 19:17:19
half are 1 more half are 2 more
rrusczyk 2020-05-14 19:17:22
Convince me.
rrusczyk 2020-05-14 19:17:34
I see a lot of you claiming "symmetry".
rrusczyk 2020-05-14 19:17:41
Convince me we have symmetry here.
rrusczyk 2020-05-14 19:18:54
It is true that the number of numbers from 1 to 900 that are 1 more than a multiple of 3 equals the number of numbers that are 2 more than a multiple of 3.
rrusczyk 2020-05-14 19:19:08
But we care here about the *product of 6 such numbers*.
rrusczyk 2020-05-14 19:19:34
Why is the product of 6 of these numbers (after we take out the multiples of 3) equally likely to be 1 more or 2 more than a multiple of 3?
gting 2020-05-14 19:19:51
If we start with 2 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 2 mod 3 or 4 mod 3, which is just 1 mod 3.

If we start with 1 mod 3, we can multiply by either 1 mod 3 or 2 mod 3 to get 1 mod 3 or 2 mod 3, which is 1 mod 3.

So out of all the 4 cases at each step, half are 1 mod 3 and half are 2 mod 3.
NinjaMango 2020-05-14 19:19:51
(3n+1)(3n+1)=9n^2+6n+1, (3n+2)(3n+1) has a 2 term at the end, and there are 2 cases, and (3n+2)(3n+2) has a four term, which is remainder 1. So two cases of each
rrusczyk 2020-05-14 19:20:14
Great -- let's make this a little more new-student friendly.
rrusczyk 2020-05-14 19:20:28
We can make a little multiplication table to see what happens when we multiply numbers that are $1$ or $2$ more than a multiple of $3$:

$$\begin{array}{c|cc} \times & 1 & 2 \\ \hline 1 & 1 & 2 \\ 2 & 2 & 1 \end{array}$$
rrusczyk 2020-05-14 19:21:02
This is just a multiplication table in which we look at the remainder of the product when we multiply two numbers.
rrusczyk 2020-05-14 19:21:26
If we multiply two numbers that are 2 more than a multiple of 3, we get a product that is 1 more than a multiple of 3. And so on.
rrusczyk 2020-05-14 19:21:36
Some of you may recognize what we're doing here as "modular arithmetic". If that term is new to you, I recommend you check it out!
rrusczyk 2020-05-14 19:21:41
The products in our table are the remainders we get when we divide the products by $3.$ We see that whether we start with a number that is $1$ or $2$ more than a multiple of $3,$ when we multiply by the next randomly-chosen number, the new product is equally likely to be $1$ or $2$ more than a multiple of $3.$
rrusczyk 2020-05-14 19:21:55
So, if the product of all $6$ numbers is not a multiple of $3,$ it is equally likely to be $1$ or $2$ more than a multiple of $3.$
rrusczyk 2020-05-14 19:22:06
That is, the probability the product is one more than a multiple of $3$ is $\frac12\cdot \frac{64}{729} = \frac{32}{729}.$
rrusczyk 2020-05-14 19:22:10
Almost there! We just have to deal with $5.$
rrusczyk 2020-05-14 19:22:13
What do we find there?
Streaks123 2020-05-14 19:22:52
Same as in this one?
cj13609517288 2020-05-14 19:22:58
It is 1/4 of the time. Same logic here except with 4 cases instead of 2.
rrusczyk 2020-05-14 19:23:05
It's basically the same deal as with $3!$ (Phew!)
Alculator11 2020-05-14 19:23:11
It can't be a multiple of 5
rrusczyk 2020-05-14 19:23:27
None of the numbers chosen can be a multiple of $5$. What's the probability that happens?
melonlord 2020-05-14 19:23:51
(4/5)^6
The_Better_Samuel 2020-05-14 19:23:51
(4/5)^6
sdattilo2002 2020-05-14 19:23:51
(4/5)^6
BakedPotato66 2020-05-14 19:23:51
$(4/5)^6$
MR_67 2020-05-14 19:23:51
(4/5)^6
FearlessTaurus 2020-05-14 19:23:57
(4/5)^6
maxben 2020-05-14 19:23:59
(4/5)^6
jai123 2020-05-14 19:23:59
(4/5)^6
rrusczyk 2020-05-14 19:24:01
None of the chosen numbers can be multiples of $5,$ which happens with probability $(4/5)^6.$
rrusczyk 2020-05-14 19:24:06
Then, we can make a multiplication table to look at possible remainders of products:

$$\begin{array}{c|cccc} \times & 1&2&3&4 \\ \hline 1 & 1&2&3&4 \\ 2 & 2&4&1&3 \\ 3 & 3&1&4&2 \\ 4 & 4&3&2&1 \end{array}$$
rrusczyk 2020-05-14 19:24:18
(You can work through that multiplication table on your own later ;) )
Jomo 2020-05-14 19:24:26
huh they are all equal...
Kruxe 2020-05-14 19:24:26
that has a cool pattern lol
MathJams 2020-05-14 19:24:29
all happen equally amounts of times
rrusczyk 2020-05-14 19:24:33
Exactly!
rrusczyk 2020-05-14 19:24:44
So...
MR_67 2020-05-14 19:24:47
1/4 that it is 4 mod 5
melonlord 2020-05-14 19:24:47
it's 1/4
MR_67 2020-05-14 19:24:56
1/4
aidni47 2020-05-14 19:24:56
1/4
sarahAops2020 2020-05-14 19:24:56
*1/4
rrusczyk 2020-05-14 19:25:14
Yep. $1/4$ of these products will end up being 4 more than a multiple of 5.
rrusczyk 2020-05-14 19:25:16
So, the probability our product is $4$ more than a multiple of $5$ is

$\frac14\cdot \frac{4^6}{5^6} = \frac{4^5}{5^6}.$
cj13609517288 2020-05-14 19:25:23
thank goodness we're done. My brain is basically ded.
rrusczyk 2020-05-14 19:25:26
Mine too.
BakedPotato66 2020-05-14 19:25:32
yay symmetry is fun!!!
rrusczyk 2020-05-14 19:25:35
Yes!
rrusczyk 2020-05-14 19:25:39
Now, how do we finish the problem?
sdattilo2002 2020-05-14 19:26:05
now we just multiply
Rsar12 2020-05-14 19:26:05
then multiply all the probabilities together
sarahAops2020 2020-05-14 19:26:05
multipy
ElNoraa 2020-05-14 19:26:05
multiply all together
MathJams 2020-05-14 19:26:05
multiply them?
fasterthanlight 2020-05-14 19:26:05
multiply the probabilities?
coldcrazylogic 2020-05-14 19:26:05
multiply everything
sixoneeight 2020-05-14 19:26:08
multiply them!
MAthen07 2020-05-14 19:26:08
Multiply em' all together.
rrusczyk 2020-05-14 19:26:12
Our three events are independent (the remainder we get upon dividing by $2,3,$ or $5$ does not depend on the other two remainders -- think later about why). So, we multiply our three probabilities.
rrusczyk 2020-05-14 19:26:14
You might want to simplify this a little bit before entering it into your calculator: $$\frac{63}{64} \cdot \frac{32}{729} \cdot \dfrac{4^5}{5^6} = \dfrac{7 \cdot 512}{81 \cdot 5^6}.$$
rrusczyk 2020-05-14 19:26:21
What do we get?
jai123 2020-05-14 19:26:36
calculator time?
rrusczyk 2020-05-14 19:26:38
For sure!
Jerry_Guo 2020-05-14 19:26:56
0.283%
MR_67 2020-05-14 19:26:56
0.283%
Kruxe 2020-05-14 19:27:11
0.283% oops
BakedPotato66 2020-05-14 19:27:11
0.283 %
prajna1225 2020-05-14 19:27:11
0.283%
onedance 2020-05-14 19:27:11
0.283 percent
acornfirst 2020-05-14 19:27:11
0.283%
bronzetruck2016 2020-05-14 19:27:11
0.283%
rrusczyk 2020-05-14 19:27:15
This computes to be $0.002831\ldots$.
rrusczyk 2020-05-14 19:27:18
As a percent to the nearest thousandth, it is $\boxed{0.283}\%$.
casp 2020-05-14 19:27:26
"rrusczyk: We're not sure! Maybe 90 minutes or so."
rrusczyk 2020-05-14 19:27:41
I was right! I think I just did 90 minutes. DPatrick did the rest
rrusczyk 2020-05-14 19:27:58
That's it for the problems tonight.
rrusczyk 2020-05-14 19:28:14
Be sure to keep playing around with the others -- there are still a bunch more to play with!
Quaoar 2020-05-14 19:28:18
This was so much fun and so enlightening! Thank you amysz, bedwards, Binomial-theorem, corinne, dcouchman, DPatrick, kdej22, LadyBlue 32, nbasrl, rponda, rrusczyk, sarahtrebatleder, tuvie, and vapodaca!
edgymemelord 2020-05-14 19:28:23
ITS BEEN 4 HOURS?!?!
rrusczyk 2020-05-14 19:28:27
No kidding!
donguri 2020-05-14 19:28:31
is this the longest ever math jam?
rrusczyk 2020-05-14 19:28:36
Probably.
TMSmathcounts737 2020-05-14 19:28:39
this lasted for 4 hours and 13 minutes.
rrusczyk 2020-05-14 19:28:52
Should have been one minute less than one hour less.
lilcritters 2020-05-14 19:28:59
Will there be a similar Math Jam next year?
rrusczyk 2020-05-14 19:29:07
Maybe, but hopefully not for the same reason!
razhder 2020-05-14 19:29:10
How many people do you think it would take to break the schoolhouse?
rrusczyk 2020-05-14 19:29:17
More than 1400, we learned today!
rrusczyk 2020-05-14 19:29:20
As a reminder, there are two more MATHCOUNTS Week events tomorrow:
rrusczyk 2020-05-14 19:29:23
Q&A with John Urschel at 3:14pm ET/12:14pm PT on the MATHCOUNTS YouTube channel.
rrusczyk 2020-05-14 19:29:25
The World's Largest Countdown Round at 6:14pm ET/3:14pm PT right back in this room. See you there!
winner320 2020-05-14 19:29:31
goodbye everyone
MokshaParam 2020-05-14 19:29:31
yay!
mmjguitar 2020-05-14 19:29:31
Thank you!!
aidni47 2020-05-14 19:29:31
Thanks so much for doing this!
Math5K 2020-05-14 19:29:50
Thank you so much!
mfro24 2020-05-14 19:29:50
Thank you so much all!
mathicorn 2020-05-14 19:29:50
I really liked this
akpi 2020-05-14 19:29:50
thank you rrusczyk
mfro24 2020-05-14 19:29:50
I can't believe it's been this long!
onedance 2020-05-14 19:29:52
will you post the transcript?
DPatrick 2020-05-14 19:29:52
Thanks for coming!
rrusczyk 2020-05-14 19:29:57
We will indeed!
rrusczyk 2020-05-14 19:30:04
Thank you everybody!
EpicGirl 2020-05-14 19:30:19
Thank You!
dolphin7 2020-05-14 19:30:19
Thank You!
mohanty 2020-05-14 19:30:19
Thank you so much!
SharonW 2020-05-14 19:30:19
THANK YOU!
Apple321 2020-05-14 19:30:19
thanks
arirah9 2020-05-14 19:30:19
Thank you so much!!!
huela 2020-05-14 19:30:19
Thank you!!
karthic7073 2020-05-14 19:30:25
Thanks for teaching!
Coolpeep 2020-05-14 19:30:25
bye! thank you!
MathWiz20 2020-05-14 19:30:25
Bye! Yay! Thank you so much!
johnnyboy1113 2020-05-14 19:30:25
Yay! Thanks a lot
Beastboss 2020-05-14 19:30:28
Thank you!
PracticingMath 2020-05-14 19:30:40
Thank you!
Shubhashubha 2020-05-14 19:30:43
Thank YOU
brainyharry 2020-05-14 19:30:45
where will you post the transcript
rrusczyk 2020-05-14 19:30:59
https://artofproblemsolving.com/school/mathjams-transcripts
ohjoshuaoh 2020-05-14 19:31:30
Thank you everyone!!!
akpi2 2020-05-14 19:31:30
TTHHAANNKKSS!!
THANKS SO MUCH
rrusczyk 2020-05-14 19:32:06
Thanks also to our superstar assistants!
Juneybug 2020-05-14 19:32:25
how do you pronounce your last name rusczyk?
rponda 2020-05-14 19:32:27
Thank you.
rrusczyk 2020-05-14 19:32:33
RUH-sick (just like it looks )
TheEpicCarrot7 2020-05-14 19:32:58
Lol imagine we ended at 6:28
mohanty 2020-05-14 19:32:58
WE STARTED AT PI TIME(EST) AND ENDED AT TAU TIME(CST)